Secondary Education Examination (SEE) Exam oriented site

UNIT 7

MOTION AND FORCE

Portrait of Isaac Newton (1642-1727)

Exercise

1. Choose the correct option for the following questions:

(a) What is the relation between the distance between two objects (d) and the gravitational force (F) produced between them?

(i) F ∝ 1/d

(ii) F ∝ d²

(iii) F ∝ 1/d²

(iv) F ∝ d

(b) What is the change in the gravitational force between two objects when their mass is doubled?

(i) the force doubles

(ii) the force becomes four-time

(iii) the force is reduced two time

(iv) the force is decreased four times

(c) If the gravitational force between two objects on Earth is 60 N. what is the gravitational force between those two objects on the moon?

(i) 10 N

(ii) 6 N

(iii) 9.8 N

(iv) 60 N

(d) Which one of the following statements is correct?

(i) The value of acceleration due to gravity increases as we go deeper from the surface of the earth.

(ii) The value of acceleration due to gravity decreases as the height above the surface of the earth increases.

(iii) The value of acceleration due to gravity is less in the polar region than that in the equatorial region.

(iv) The value of the acceleration of gravity is highest at the highest place on the Earth.

(e) At which of the following places do you weigh the most?

(i) peak of Mount Everest

(ii) peak of Api Himal

(iii) Kechnakwal of Jhapa

(iv) Chandragiri Hills

(f) The radius of the Earth is 6371 km and the weight of an object on the earth is 800 N. What is the weight of the object at a height of 6371 km from the surface of the earth?

(i) 800N

(ii) 1600 N

(iii) 200 N

(iv) 3200 N

(g) If the mass and the radius of a celestial body are two times the mass and the radius of the earth respectively, what is the value of acceleration due to the gravity of that body?

(i) 9.8 ms

(ii) 4.9 m/s²

(iii) 19.6 m/s²

(iv) 10 m/s²

(h) What will be the weight of a man on the moon, if his weight on earth is 750 N? (The acceleration due to the gravity of the moon = 1.63 m/s² )

(i) 124.74N

(ii) 125 N

(iii) 126.8 N

(iv) 127.8 N

(i) The mass of planet B is twice the mass of planet A but its radius is half of the radius of planet A. Similarly, the mass of planet C is half of the mass of planet A. but its radius is twice the radius of planet A. If the weight of an object in planets A, B and C is W1, W2 and W3 respectively, which of the following order is correct?

(i) W2 > W3 > W2

(ii) W2 > W1 > W3

(iii) W1 > W2 > W3

(iv) W2 > W3 > W1

(j) Which one of the following conclusions is correct while observing a freely falling object every second?

(i) the distance covered increases uniformly

(ii) velocity increases uniformly

(iii) acceleration increases uniformly

(iv) translation takes place uniformly

2. Differentiate between:

(a) Gravitational constant G and acceleration due to gravity g

Difference Between Gravitational Constant (G) and Acceleration Due to Gravity (g)

Gravitational Constant (G)	Acceleration Due to Gravity (g)
1. The gravitational constant (G) is a universal constant that appears in Newton's Law of Gravitation.	1. The acceleration due to gravity (g) is the acceleration experienced by a mass due to the Earth's gravitational field.
2. The value of G is constant throughout the universe, and its value is approximately 6.67 × 10-11 Nm2/kg2.	2. The value of g varies from place to place on the Earth's surface. On the surface of the Earth, it is approximately 9.8 m/s2.
3. G is a fundamental constant of nature.	3. g is dependent on the mass of the Earth and the distance from the Earth's center.
4. The value of G does not change with the location on Earth.	4. The value of g decreases with altitude and increases with depth under Earth's surface.
5. Gravitational constant is used to calculate the force between two point masses using the equation F = Gm1m2  / d2.	5. Acceleration due to gravity (g) is used to calculate the weight of an object using the equation W = mg, where W is weight and m is the mass of the object.

(b) Mass and Weight

Difference Between Mass and Weight

Mass	Weight
1. Mass is a measure of the amount of matter in an object.	1. Weight is the force exerted by gravity on an object.
2. Mass is a scalar quantity and does not have a direction.	2. Weight is a vector quantity as it has both magnitude and direction (direction towards the center of the Earth or any other gravitational source).
3. The mass of an object is constant and does not change with location.	3. The weight of an object varies depending on the gravitational field of the body where the object is located (e.g., on the Moon, weight is one-sixth of that on Earth).
4. The SI unit of mass is kilogram (kg).	4. The SI unit of weight is Newton (N), which is a unit of force.
5. Example: A person has a mass of 70 kg everywhere, regardless of location.	5. Example: The same person weighs about 686 N on Earth but will weigh only 114 N on the Moon.

3. Give reason:

(a) Acceleration due to gravity is not the same in all parts of the earth.

→ We know acceleration due to gravity is inversely proportional to the square of radius of earth, i.e. g ∝ 1/R² and earth is not a perfectly spherical due to which its radius varies from place to place. Hence, the acceleration due to gravity is not the same in all parts of earth.

(b) Jumping from a significant height may cause more injury.

→ If a person jumps from certain height, his acceleration goes on increasing towards the earth surface. Since, force is directly proportional to acceleration due to gravity (F ∝ g), the increase in acceleration causes increase on force with which he strikes on surface of earth. Hence, jumping from a significant height may cause more injury.

(c) Mass of Jupiter is about 319 times the mass of the Earth, but its acceleration due to gravity is only about 2.6 times the acceleration due to gravity of the Earth.

→ Mass of Jupiter is about 319 times the mass of the Earth, but its acceleration due to gravity is only about 2.6 times the acceleration due to gravity of the Earth because acceleration due to gravity depends not only on the mass but also on the radius of the planet, i.e. g = GM/R². The radius of Jupiter is much greater than that of earth. Thus the acceleration due to gravity is not large even mass of Jupiter is 319 times more than that of earth.

(d) Among the objects dropped from the same height in the polar region and the equatorial region of the earth, the object dropped in the polar region falls faster.

→ We know acceleration due to gravity is inversely proportional to the square of radius of earth i.e. g ∝ 1/R² and polar radius of earth is less than equatorial radius of earth as earth is not perfectly spherical. Hence, the less radius in polar region causes more acceleration to act upon object falling from polar region. Thus the object dropped in polar region falls faster.

(e) Out of two paper sheets, one is folded to form a ball. If the paper ball and the sheet of paper are dropped simultaneously in the air, the folded paper will fall faster.

→ If the paper ball and the sheet of paper are dropped from certain height, paper ball falls faster because larger the surface area larger will be the air resistance and vice versa. So, the velocity of paper ball increases due to decrease in air resistance.

(f) When a marble and a feather are dropped simultaneously in a vacuum, they reach the ground together (at the same time).

→ In vacuum, there is no air resistance. Hence a marble and feather dropped simultaneously in vacuum reach in ground together at same time as they have same velocity.

(g) As you climb Mount Everest, the weight of the goods that you carry decreases.

→ The value of ‘g’ is inversely proportional to the square of the height from earth’s surface. So, the value of ‘g’ goes on decreasing while we move towards the Mount Everest. Since, weight is directly proportional to the value of ‘g’. Hence, the weight of goods also decreases as we climb Mount Everest.

(h) It is difficult to lift a big stone on the surface of the earth. but it is easy to lift a smaller one.

→ The acceleration due to gravity is equal on both sized stone and weight directly depends on mass of object taken. Big stone has large weight which causes large force of gravity towards the center of earth. Thus we should apply large force to oppose force of gravity for big stone. But small stone has less weight due to which it experiences less force of gravity and less force is required to oppose its force of gravity. Hence, it is easy to lift small stone than big one on earth’s surface.

(i) Mass of an object remains constant but its weight varies from place to place.

→ Mass is amount of matter contained in a body. Since, the amount of matter doesn’t change from place to place, mass of an object remains constant.

On the other hand, Weight of an object is the gravitational force between earth and that object. So, Weight (W) = GMm/R². Since, earth is not perfectly spherical, so its radius is different at different place due to which the weight also varies from place to place.

(j) One will have an eerie feeling when he/she moves down while playing a Rote Ping.

→ One will have an eerie feeling when he/she moves down while playing a Rote Ping due to the sensation of free fall, where the person experiences temporary weightlessness. This occurs because, during free fall, both the person and the surroundings accelerate downward at the same rate due to gravity, making it feel as though there is no normal force acting on the body.

4. Answer the following questions:

(a) What is gravity?

→ Gravity is defined as the force of attraction with which the earth like massive objects pulls other lighter objects towards its center.

(b) State Newton's universal law of gravitation.

→ Newton’s universal law of gravitation states that, the force of attraction between two heavenly bodies is directly proportional to the product of their masses and inversely proportional to the square of distance between their center.

Mathematically, F = Gm₁m₂ / d² , Where F is force of attraction between two bodies, m₁ and m₂ are masses of two different bodies, d is the distance between two bodies from their center and G is constant called universal gravitational constant whose value is 6.67 x 10^-11 Nm²/Kg².

(c) Write the nature of gravitational force.

→ Nature of gravitational force:

· Gravitational force is directly proportional to the magnitude of product of masses of two heavenly bodies.

· Gravitational force is inversely proportional to the square of distance between two heavenly bodies from their center.

· Gravitational force is weakest among the four fundamental force. [Learn more about four fundamental force here]

· It is always attractive and never repulsive.

· It is long range force.

(d) Define gravitational constant (G).

→ The gravitational constant (G) is defined as the magnitude of gravitational force produced between two unit masses that are separated by unit distance.

(e) Under what conditions is the value of gravitational force equal to the gravitational constant (F=G)?

→ Under any of following conditions the value of F is equal to G:

· If two heavenly bodies have unit mass and are separated by unit distance.

· If the product of masses of two heavenly bodies is equal to the square of distance from their center.

(f) Write two effects of gravitational force.

→ Two effects of gravitational force are as follows:

· The existence of solar system is due to the gravitational force acting anong the sun, various planets, satellites and other bodies in the system.

· Tides are common on ocean due to the force of gravitation between moon and sun.

(g) Mathematically present the difference in the gravitational force between two objects when the mass of each is made double and the distance between them is made one forth their initial distance.

→ Let,  

$m_1$ and $m_2$the initial masses​, and the initial distance be 

$r$. The initial gravitational force $F_{\text{initial}}$ is given by:

$$F_{\text{initial}} = \frac{G \cdot m_1 \cdot m_2}{r^2}$$

Now, the masses are doubled, so the new masses are $2m_1$ and $2m_2$, and the distance is reduced to one-fourth, so the new distance is $\frac{r}{4}$. The new gravitational force $F_{\text{final}}$ is:

$$F_{\text{final}} = \frac{G \cdot (2m_1) \cdot (2m_2)}{\left( \frac{r}{4} \right)^2}$$

Simplifying the equation:

$$F_{\text{final}} = \frac{4 \cdot G \cdot m_1 \cdot m_2}{\frac{r^2}{16}} = 64 \cdot \frac{G \cdot m_1 \cdot m_2}{r^2}$$

Thus,

$$F_{\text{final}} = 64 \cdot F_{\text{initial}}$$

Therefore, the final gravitational force is 64 times the initial gravitational force.

(h) What is gravitational force?

→ The mutual force of attraction between any two objects in universe is called gravitational force.

(i) Define acceleration due to gravity.

→ The acceleration produced on a lighter object due to gravitational attraction of a massive object like planets or satellites is called acceleration due to gravity.

(j) What is free fall? Give two examples of it.

→ The condition in which any object falls without any external resistance is called free fall. Example: falling of objects on moon surface, large mass falling from small height in earth’s surface

(k) Under what conditions is an object said to be in free fall?

→ Under following condition, an object is said to be in free fall:

· When any object falls without any external resistance.

OR

· When an object falls under the effect of gravity alone.

(l) Write the conclusions of the feather and coin experiment.

→ The feather and coin experiment confirms that acceleration of freely falling bodies, (whatever be their masses) is the same in the absence of air. But in presence of air, the object with larger surface area has more resistance.

(m) What is weightlessness?

→ The weightless condition of an object during freefall is called weightlessness.

(n) Mention any four effects of gravitational force.

→ Effects of gravitational force are as follows:

· The existence of solar system is due to the gravitational force acting along the sun, various planets, satellites and other bodies in the system.

· Tides are common on ocean due to the force of gravitation between moon and sun.

· The weight of objects is the result of gravitational pull acting on their mass.

· It binds stars, gas and dust together forming galaxies.

(o) Prove that acceleration due to the gravity of the Earth is inversely proportional to the square of its radius (g ∝ 1 / R)

→ Newton's law of universal gravitation states that the gravitational force (

$F$) between two masses $M_1$ and $M_2$ is given by:

$$F = G \cdot \frac{M_1 M_2}{r^2}$$

Where:

• $F$ is the force between the two masses,
• $G$ is the gravitational constant ($6.67\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2$
• $\mathrm{r\; is\; the\; distance\; between\; the\; centers\; of\; the\; two\; masses.}$

Let's consider an object of mass $m$ placed on or near the surface of a earth (with mass $M$ and radius $R$). The force of gravity acting on the object due to the planet will be the gravitational force between the planet and the object.

In this case, we have:

• $M_1 = M$ (mass of the planet),
• $M_2 = m$ (mass of the object),
• $r = R$e distance from the center of the planet to the object).

Thus, the gravitational force acting on the object is:

$$F=G\cdot \frac{M\cdot m}{R^{\mathrm{2}}}$$

Since this is the force due to gravity acting on the object.

According to Newton's Second Law of Motion:

$$F=m\cdot \mathrm{g \; [F\; is\; the\; gravitational\; force,\; m\; is\; the\; mass\; of\; the\; object,\; g\; is\; the\; acceleration\; due\; to\; gravity\; gravity.}$$

Since both expressions represent the force on the object, we can set them equal to each other:

$$m\cdot g=G\cdot \frac{M\cdot m}{R^{\mathrm{2}}}$$

$$g = G \cdot \frac{M}{R^2}$$

This is the desired formula for the acceleration due to gravity at the surface of the planet. It tells us that:

$$g = \frac{GM}{R^2}$$​

                                                           or,     g ∝ 1 /  R2            (If G and M are taken as constant)

This equation expresses that the acceleration due to gravity 'g' is inversely proportional to the square of the radius of the planet (R).

(p) Mention the factors that influence acceleration due to gravity.

→ The factors which influence acceleration due to gravity are:

· Mass of heavenly body

· Radius of heavenly body

· Height from surface of heavenly body

· Depth from surface of heavenly body

(q) The acceleration due to the gravity in the Earth surface is 9.8 m/s². What does this mean?

→ It means that the velocity of a body rises by 9.8 m/s in each seconds during its fall if air resistance is neglected.

(r) Mass of the Moon is about 1/81 times the mass of the Earth and radius of earth is about 3.7 times that of moon. If the earth is squeezed to the size of the moon, what will be the effect on its acceleration due to gravity? Explain with the help of mathematical calculation.

(s) The acceleration due to gravity of an object of mass 1 kg in outer space is 2m/s². What is the acceleration due to gravity of another object of mass 10 kg at the same point? Justify with arguments.

→ The acceleration due to gravity of the second object, which has a mass of 10 kg, will still be 2 m/s² at the same point in outer space.

Justification:

· Formula for acceleration due to gravity is g = GM / r², where g depends solely on the mass of the gravitating body (M) and the distance (r).

· As both objects are located at the same point in space, the values of M and r do not change.

Hence, the 10 kg object will also experience an acceleration due to gravity of 2 m/s².

(t) A man first measures the mass and weight of an object in the mountain and then in the Terai. Compare the data that he obtains.

→ Comparison of his data:

· The mass of the object stays the same whether it's on the mountain or in the Terai, since mass does not change with location.

· The weight of that object will be higher in the Terai because acceleration due to gravity (g) is a bit stronger at lower altitudes than it is on the mountain.

(u) A student suggests a trick for gaining profit in a business. He suggests buying oranges from the mountain selling them to Terai at the cost price. If a beam balance is used during this transaction, explain, based on scientific fact, whether his trick goes wrong or right.

→ His trick of buying oranges from mountain and selling it at Terai on cost price goes wrong because beam balance measures mass of oranges, not weight. Since the mass of oranges remains same both at mountain and terai. So, there will be no profit in his business. (Since, there will be transaction cost of oranges, so he will be at loss)

(v) How is it possible to have a safe landing while jumping from a flying airplane using a parachute? Is it possible to have a safe landing on the moon in the same way? Explain with reasons.

→ It is safe to jump from a flying airplane using a parachute on earth because the parachute has large surface area due to which it experiences large air resistance. Hence, the person falling from the parachute will experience less acceleration and have a safe landing on earth.

However, if the same activity is performed on moon, the person is likely to get injured as moon has no atmosphere. Despite the gravity of the moon being less, falling from an airplane with a parachute causes it to gain extreme speed without air resistance. So, it is not safe to jump from a flying airplane in the moon.

(w) The acceleration of an object moving on the earth is inversely proportional to the mass of the object, but for an object falling towards the surface of the earth, the acceleration does not depend on the mass of the object, why?

→ When an object is moved, the acceleration is determined by the applied force and the mass of that object (∵ F=ma → a = F/m). For the same force, a larger mass resists the motion due to inertia, resulting in less acceleration.

However, when an object is falling towards surface of earth, the force acting on object’s weight will be F = mg. Hence the acceleration a = F/m → a = mg/m → a = g. This means acceleration is independent of mass.

5. Solve the following mathematical problems: [Note: For best view of numericals open the page in this way]

(a) The masses of two objects A and B are 20 kg and 40 kg respectively. If the distance between their centers is 5 m, calculate the gravitational force produced between them.

→ Solution:

     Mass of object A = m₁  = 20 kg

     Mass of object B = m₂ = 40 kg

    Distance between their centers (d) = 5 m

     Now, 

             Gravitational force is given by

     F = Gm₁m₂ / d² 

          = (6.67 × 10^-11) × (20 × 40) / 5²  

          = (6.674 × 10^-11) × 800 / 25

          = (6.674 × 10^-11) × 32

          = 2.14 × 10^-9 N

(b) Calculate the gravitational force between the two bodies shown in the figure.

→ There's no any figure given in book without which calculation is not possible.

(c) Mass of the Sun and Jupiter are 2 × 10³⁰ kg and 1.9 × 10²⁷ kg respectively. If the distance between Sun and Jupiter is 1.8 × 10⁸ km, calculate the gravitational force between Sun and Jupiter.

→ Solution:

     Mass of sun = m₁ = 2 × 10³⁰ kg

     Mass of Jupiter = m₂ = 1.9 × 10²⁷ kg

     Distance between their centers = d = 1.8 × 10⁸ km = 1.8 × 10¹¹ m        

     Now, 

             Gravitational force is given by 

    F = Gm₁m₂ / d²

       = (6.67 × 10^-11) × (2 × 10³⁰ × 1.9 × 10²⁷) / (1.8 × 10¹¹)²

       = (6.67 × 10^-11) × (3.8 × 10⁵⁷) / 3.24 × 10²²

         F ≈ 7.82 × 10²⁴ N        

(d) Gravitational force produced between the Earth and Moon is 2.01 × 10²⁰ N If the distance between these two masses is 3.84 × 10⁵ km and the mass of the earth is 5. 972 × 10²⁴ kg, calculate the mass of the moon.

→ Solution:

     Mass of earth = m₁  = 5. 972 × 10²⁴ kg

     Mass of moon = m₂ = ?

     Gravitational force = F = 2.01 × 10²⁰ N

     Distance between their centers (d) = 3.84 × 10⁵ km = 3.84 × 10⁸ m        

     Now, 

            Gravitational force = F = Gm₁m₂ / d²

            or, m₂ = Fd² / Gm₁

                       = 2.01 × 10²⁰ (3.84 × 10⁶)² / (6.67 × 10^-11).(5. 972 × 10²⁴) 

                       = 7.67 × 10²²  kg.

(e) Gravitational force produced between the Earth and the Sun is 3.54 × 10²² N If the masses of the Earth and sun are 5.972 × 10²⁴ kg and 2 × 10³⁰ kg respectively, what is the distance between them?

→ Solution:

     Mass of earth = m₁  = 5.972 × 10²⁴ kg

     Mass of sun = m₂ = 2 × 10³⁰ kg

     Gravitational force = F = 3.54 × 10²² N

     Distance between their centers = d = ?        

     Now, 

           Gravitational force = F = Gm₁m₂ / d² 

                           = (6.67 × 10^-11).(5.972 × 10²⁴).(2 × 10³⁰) / (3.54 × 10²²)

                        d² = 2.25 × 10²²

                  or, d = 1.5 × 10¹¹ m

(f) The mass of the moon is 7.342 × 10²² kg. If the average distance between the earth and the moon is 384400 km, calculate the gravitational force exerted by the moon on every kilogram of water on the surface of the earth.

→ Solution:

     Mass of water on earth = m₁  = 1 kg

     Mass of moon = m₂ = 7.342 × 10²²

     Gravitational force (F) = ?

     Distance between their centers (d) = 384400 km = 384400000 m        

     Now, 

            Gravitational force = F = Gm₁m₂ / d²

            or, F = (6.67 × 10^-11) . (1) . (7.342 × 10²²) / (384400000)²

                    = 3.314 × 10^-5 N

(g) If the mass of the moon is 7.342 × 10²² kg and its radius is 1737 km. calculate its acceleration due to gravity.

→ Solution:

    Mass of Moon (M) = 7.342 × 10²²

    Radius (r) = 1737 km = 1737000 m

    Acceleration due to gravity (g) = ?

    We know that,

            g = GM / r²

       or, g = (6.67 × 10^-11) . (7.342 × 10²²) / (1737000)²

               = 1.623 m/s²

(h) Mass of the Earth is 5.972 × 10²⁴ kg and the diameter of the moon is 3474 km. If the earth is compressed to the size of the moon, how many times will be the change in acceleration due to the gravity of the earth so formed than that of the real Earth?

→ Solution:

    Mass of Earth (M) = 5.972 × 10²⁴ kg [it remains unchanged]

    We know the diameter of earth is 12742 km. 

    So, Original radius of Earth = R_o = 12742/2 = 6371 km = 6371000 m 

    Diameter of Moon = 3434 km = 3474000 m

    So, Radius of Moon = 3474000/2 = 1737000 m

According to question, 

    radius of moon = Compressed radius of Earth = R_c = 1737000 m 

Now, 

        Acceleration due to gravity (g) = GM / R²

        Since, the mass of earth remains same before and after compression,

        ratio of acceleration due to gravity before and after compression is given by

                   g_c / g_o = (R_o)² / (R_c)²

             or, g_c  / g_o = (6371000)² / (1737000)²

             ∴ g_c  / g_o = 13.45

Hence, new acceleration due to gravity of earth will be 13.45 times the acceleration due to gravity of earth before compression.

 

(i) If the mass of Mars is 6.4 × 10²³ kg and its radius is 3389 km, calculate its acceleration due to gravity. What is the weight of an object of mass 200 kg on the surface of Mars?

→ Solution:

   Mass of Mars (M) = 6.4 × 10²³ kg

   Radius of mars (r) = 3389 km = 3389000 m

   We know,

   Acceleration due to gravity (g)   = GM/r²

                                                       = 6.67 × 10^-11 × 6.4 × 10²³ / (3389000)2

                                                       = 42.688 × 10¹² / (3.389 × 10⁶)²

                                                       = 42.688 / 1.1485 × 10

                                                       = 3.716 m/s2

   Again,

   Mass = 200 kg, Weight (W) = ?

   We know, Weight (W) = Mass × gravity

                                       = 200 × 3.716 

                                       = 743.2 N

(j) The acceleration due to the gravity of the earth is 9.8 m/s². If the mass of Jupiter is 319 times the mass of the Earth and its radius is 11 times the radius of the Earth, calculate the acceleration of gravity of Jupiter. What is the weight of an object of mass 100 kg on Jupiter?

→ Solution:

   Earth’s acceleration due to the gravity (g) = 9.8 m/s²

   Mass of Earth = m (say)

   Mass of Jupiter = 319m

   Radius of earth = r (say)

   Radius of Jupiter = 11r

   Acceleration due to gravity at Jupiter = g_j (say)

   Now,

                For earth: g = Gm/r² → 9.8 = Gm/r² ……. (i)

                For Jupiter: g_j = G × 319m /(11r)² ………. (ii)

   Dividing equation (ii) by (i), we get: g_j / 9.8 = 319/121 → g_j = 25.83 m/s

   Again,

            Weight of object of mass 100 kg (W) = mass × gravity

                                                                        = 100 × 25.83 = 2583 N

(k) Earth's mass is 5.972×10²⁴ kg and its radius is 6371 km. Calculate the acceleration due to the gravity of the earth at the height of the artificial satellite shown in the figure.

→ There's no any figure given in book without which calculation is not possible.

(l) Mass of the earth is 5.972×10²⁴ kg and its radius is 6371 km. If the height of Mt. Everest is 8848.86 m from the sea level, calculate the weight of an object of mass 10 kg at the peak of Mt. Everest.

→ Solution:

   F = gravitational force (weight)

   mass of the Earth (M) = 5.972×10²⁴ kg

   mass of the object (m) = 10 kg

   Earth's radius = R = 6371 km = 6371000 m

   Add the height of Mt. Everest (h) = 8848.86 m

  distance from the Earth's center to the object = d = R+h 

                                                                                = 6371000 + 8848.86 = 6379848.86 m 

                                        [where R is the Earth's radius and h is the height of Mt. Everest.]

    Now,

    We know,

    Gravitational force = F = GMm / d²

                = (6.67 × 10^-11) (5.972×10²⁴ )×10 / (6379848.86)²

                = 97.93 N

    Thus, the weight of the 10 kg object at the peak of Mt. Everest is approximately 97.93 N.

(m) The acceleration due to gravity of the Mars is 3.75m/s². How much mass can a weight-lifter lift on Mars who can lift 100 kg mass on the Earth?

→ Solution:

   To calculate how much mass a weight-lifter can lift on Mars, we use the relationship:

$$\mathrm{ \; W}=m\cdot \mathrm{g\; [}$$Where, W= weight , m= mass and g= acceleration due to gravity]

   Let us first determine his weight-lifting capability on Earth

  The weight-lifter can lift a mass of $100\text{<i>\hspace{0.17em}</i>}\text{<i>kg </i>}$on Earth. 

  On Earth, the acceleration due to gravity is ${\mathrm{g }}_{}=$9.8 m/s2

  The corresponding weight is:

$${\mathrm{ \; W}}_{}=m_{}\cdot {\mathrm{g }}_{}=\mathrm{<i>100 </i>}$$× 9.8=980 N

  Now, let us relate the same weight to Mars gravity:

  On Mars, the acceleration due to gravity is 3.75 m/s2

  Using the formula $W=\mathrm{m\text{'}}\cdot \mathrm{g\text{'}<span\; style="font-weight:\; normal;">, we\; can\; solve\; for</span>}$${\mathrm{ mass\; he\; can\; lift\; at\; mars}}_{}$:

$${\mathrm{ \; m}}_{}$$' $$=\frac{{\mathrm{W\; / }}_{}}{g_{}}$$'$$\frac{{}_{}}{}=<span\; style="font-weight:\; normal;"><i> </i></span>\frac{\mathrm{<i>980\; /</i>}}{}$$ 3.75 = 261.33 kg

  The weight-lifter who can lift $100\text{kg}$ on Earth will be able to lift approximately 261.33 kg on    Mars.

(n) When a stone is dropped from a bridge over a river into the water, after 2.5 seconds the sound of the stone hitting the surface of the water is heard. Calculate the height of the bridge from the surface of the water. (g = 9.8 m/s² )

→ Solution:

    Initial velocity of stone (u) = 0 m/s

    Time taken by stone to hit water surface (t) = 2.5 s

    Acceleration of stone (g) = 9.8 m/s2

    The height of bridge above the surface of water (h) = ut + 1/2 (gt2)

                                                                                       = 0 + 1/2 (9.8 × 2.5 × 2.5)

                                                                                       = 30.625 m

    Hence the height of bridge above surface of water is 30.625 meters.

(o) If a stone is dropped from a height of 15 m. how long will it take to reach the ground? Calculate the velocity of the stone when it hits the ground. 

→ Solution:

    height from which stone is dropped (h) = 15 m

    initial velocity (u) = 0 m/s

    time taken to reach ground (t) = ?

    velocity of stone when it hits the ground (v) = ?

    Now, 

            For object in free fall, h = ut + 1/2 (gt2)

                                            or, 15 = 0 + 1/2 (9.8 × t2)

                                            or, 15 / 4.9 = t2

 

                                            or, t = 1.749 s

    Also, 

            For final velocity of stone, we use (v) = u + gt

                                                                         = 0 + 9.8 × 1.749 = 17.14 m/s

    Hence, the stone will take 1.749 s to reach the ground and its velocity will be 17.14 m/s when     it hits the ground.

(p) If a cricket ball is thrown vertically upwards into the sky with a velocity of 15 m/s, to what maximum height will the ball reach?

→ Solution:

    Intial velocity (u) = 15 m/s

    acceleration of the ball (g) = -9.8 m/s2

    maximum height to which ball reach (h) =?

    Now, we know, v = u + gt

                            or, 0 = 15 + (-9.8) t [Since, velocity at maximum height is 0, so v =0]

                            or, t = 1.53 s

    Again, we know, h = ut + 1/2 gt2

                           or, h = 15 × 1.53 + 1/2 (-9.8 × 1.53 × 1.53)

                           or, h = 22.95 - 1/2 × 22.94

                                h = 22.95 - 11.47 = 11.48 m

    Hence, The ball will reach height of 11.48 m in maximum.

 

QUESTIONS ASKED IN SEE

1. What change ocuurs in the gravitational force between two bodies when their masses are kept constant and the distance between their centres is reduced by 3 times? [SEE 2080 Koshi]

→ If the distance between two bodies is reduced by three times while keeping their masses constant, the gravitational force between them increases by 9 times.

2. Prove that unit of gravitational constant is derived unit. [SEE 2080 Koshi]

→ We know,  From Newtons universal law of gravitational:

        Gravitational force = F = Gm₁m₂ / d²

    or, G = Fd² / m₁m₂

From unit wise analysis, 

    unit of G = Kg . m . s^-2 . m / Kg. Kg

                    = Kg^-1m²s^-2

Since, unit of gravitational constant depends upon three fundamental units Kg, m, and s. So, unit of Gravitational constant is derived unit.

3. A man found 90 N weight of an object of 9 kg mass at a place. From the same place the object was dropped down for 2 s to reach on the ground assuming air resistance is zero. [SEE 2080 Koshi]

i) From what height did the object drop?

→ Here,     

        Weight (W) = 90 N        height (h) = ?

        Mass (M) = 9 kg

        time (t) = 2 s

We know, W = Mg [for falling object]

               or, g = W/M

               or, g = 90 / 9 = 10 m/s²

Now, h = ut + 1/2 gt²

      or, h = 1/2 × (10) × (2)²

                =20 m

Hence, the object was dropped from 20 m height.

ii) If a stone of 500 g is dropped down from the same heights, how long does it take to reach the ground? Explain with reason.

→ The stone will take 2 seconds to reach the ground. This is because in free fall, the time taken depends only on the height and gravitational acceleration, and not on the mass of the object.

4. Which formula will you use to find out height from the earth surface from where a stone is dropped down and time taken by the stone to reach the ground is known? [SEE 2080 Madhesh]

→ To find the height from which a stone is dropped when the time taken to reach the ground is known, the formula used is derived from the equation of motion for free fall:

    h = 1/2 gt²

 

5. Rita and Pema dropped stone from the top of two different towers standing nearby. Rita dropped the stone of mass 1 kg. After 1 s time Pema dropped the stone of mass 1.5 kg which reached the ground 0.46 s later than the Rita's stone. If the tower from where Rita dropped the stone is 60 m high, what is the height of tower from where Pema dropped the stone? [SEE 2080 Madhesh]

→ Here,

       Rita's Case: 

    Mass of stone (m₁) = 1 kg

    Height of tower (h₁) = 60 m

    Time taken by it to reach ground (t₁) = ?

We know, h = 1/2 g(t₁)²

                or, 60 = 1/2 × 10 × (t₁)²

                or, (t₁)² = 12

                t₁ = 3.46 s

    Pema's' Case:

    Mass of stone (m₂) = 1.5 kg

    Height of tower (h₂) = ?

    Time taken to reach ground (t₂) = 0.46 + t₁ - 1 

                                                        = 0.46 + 3.46 - 1

                                                        = 2.92 s

Now, using same formula for free fall:

                 h₂ = 1/2 g(t₂)²

                      = 1/2 × 10 × (2.92)²

                      = 5 × 8.5264

                      = 42. 632 m

Hence,  heigth of tower from where pema dropped stone was 42. 632 m from ground.

6. What is weight of a body having 50 kg mass on the moon? [SEE 2080 Bagamati]

    i) 8.35 N    

    ii) 83.5 N     

    iii) 835 N     

    iv) 8355 N

7. Acceleration due to gravity of three heavenly bodies are given in the table alongside. On the basis of it, write the answer of following. [SEE 2080 Bagamati]

Heavenly Body      →   |       A              |              B                |            C         

Acceleration (g)     →   |  25 m/s2        |          9.8 m/s2         |       1.67 m/s2  

i. In which heavenly body it is easier to lift object? why?

→ It is easier to lift object in heavenly body 'C' because the ease of lifting an object depends on the gravitational force acting on it, which is determined by the acceleration due to gravity (

$g$) at the surface of heavenly body. The smaller the $g$, the easier it is to lift an object.

ii. If an object is dropped from the same height in all three heavenly bodies, in which heavenly body does the object fall faster? why? 

→ The object falls faster in 'A' because the rate at which an object falls depends on the acceleration due to gravity (g) of the heavenly body. Higher value of '$\mathrm{g\text{'}}$ means the object accelerates faster and thus falls faster.

8. Keeping the mass of two bodies constant, the distance between them is doubled. What will be the gravitational force exerted between them? [SEE 2080 Gandaki, SEE 2080 Grade Increment exam]

    i. F = Gm₁m₂ / d²

    ii. F = Gm₁m₂ / 2d²

    iii. F = 2Gm₁m₂ / d²

    iv. F = Gm₁m₂ / 4d²

9. In the given condition what will be the gravitational force produced between P and Q and between Q and R? In which case gravitaitonal force is more by how much? calculate. 

[SEE 2080 Gandaki]

Q No. 23

→ Here,

    CASE I - Force between P and Q

     mass of P = m₁ = 500 kg

     mass of Q = m₂ = 1100 kg

     distance between P and Q = d = 600 m

    We know, 

            F = Gm₁m₂ / d²

                = (6.67 × 10^-11) (500 × 1100) / (600)²

                =  (6.67 × 10^-11) (550000) / (360000)

                = 1.019 × 10^-10 N

Thus Gravitational force between P and Q is 1.019 × 10^-10 N.

CASE II - Force between Q and R

    mass of Q = m₂ = 1100 kg

    mass of R = m₃ = 900 kg

    distance between Q and R = d₃ = 300 + 100 = 400 m

Now, 

    Force beteen Q and R is  

             F = Gm₂m₃ / (d₃)²

                = (6.67 × 10^-11) (1100 × 900) / (400)²

                =  (6.67 × 10^-11) (990000) / (160000)

                = 41.27 × 10^-11 N

                = 4.127 × 10^-10 N

Thus, Gravitational force between Q and R is 41.27 × 10^-11 N

Hence, gravitational force between Q and R is more stronger than force between P and Q by 3.108×10^-10 N.

10. What is the reason of increasing acceleration due to gravity, if the earth is compressed to the size of the moon. [SEE 2080 Lumbini]

    i. by increasing radius of earth

    ii. by decreasing radius of earth

    iii. by increasing mass of earth

    iv. by decreasing mass of earth

11. Mass of objects A and B as shown in diagram are 2500 kg and 4000 kg respectively. [SEE 2080 Lumbini]

Q No. 23

i. calculate the gravitational force between the objects A and B.

→ Here, 

    mass of A = m₁ = 2500 kg

    mass of B = m₂ = 4000 kg

    distance between them = d = 600 m

    

    We know,

    Gravitational force = F = Gm₁m₂ / d²

                = (6.67 × 10^-11) . (2500 × 4000) / (600)²

                = 185.27 × 10^-11 N

                = 1.8527 × 10^-9 N

ii. If acceleration due to gravity on body 'A' is 9.8 m/s²  when it is on the surface of the earth. At which height from the surface of earth the body 'A' should be taken so that acceleration due to gravity is 4 m/s²  ? 

→ Solution:

    Acceleration due to gravity at surface of A is:

     g = 9.8 m/s² 

    or, GM / R² = 9.8  ........(1)

    Let that body is taken at height 'h' from surface, 

    where acceleration due to gravity = g^' = 4 m/s² 

                                    or, GM / (R+h)² = 4 ......... (2)

    Dividing equatioin (1) by equation (2):

   [GM / R²] / [GM / (R+h)²]  = 9.8 / 4 

   (R+h)² / R² = 2.45

    Taking square root on both sides

    (R+h) / R = 2.45 

    or, 1 + (h / R) = 1.565

    or, h / R = 1.565 -1 

    or, h / R = 0.565

    or, h = 0.565 × R

    or, h = 0.565 × 6371000 [∵ Radius of earth is 6371 km = 6371000 m]

    or, h = 3601192.35 m

      ∴ h = 3601.19235 km

12. What is the weight of body on moon surface if its weight is 150 N on the earth surface? [SEE 2080 Karnali]

    i. 20 N    

    ii. 25 N    

    iii. 30 N     

    iv. 35 N

13. 'Birth and death of stars was not possible in the absence of gravitational force'. Justify the statement in two points. [SEE 2080 Karnali]

→ The birth and death of stars isn't possible without gravitational force. Following points justify     this statement:

• Birth of Stars: Stars are born when gravity pulls together clouds of gas and dust in space. As these particles come closer, the temperature and pressure increase until nuclear fusion starts, forming a star. Without gravity, this process would not occur.
• Death of Stars: When a star runs out of fuel, gravity causes it to collapse. Depending on the mass of the star, it might become a white dwarf, neutron star, or black hole. This process would not be possible without the force of gravity.

14. In what velocity a stone should be thrown upward to achieve the height of 80 m from earth surface in 10 seconds. If acceleration due to gravity on earth surface is 10 m/s² at which height from the earth surface acceleration due to gravity will be 6.5 m/s²? (Given that M= 6 × 10²⁴ kg and R = 6.4 × 10⁶ m) [SEE 2080 Karnali]

→    Follow same process as in Q No 9 (ii) for solution

15. What is weight of a body having mass 100 kg on the moon? [SEE 2080 Sudurpaschim]

    i. 2500 N    

    ii. 1000 N     

    iii. 980 N    

    iv. 167 N

16. 'Role of gravitational force is very important for the birth and death of a star." Justify this statement. [SEE 2080 Sudurpaschim]

→ Same as question no 13

17. The acceleration due to gravity of three heavenly bodies are given in the table alongside. [SEE 2080 Sudurpaschim]

Heavenly Bodies                Acceleration due to gravity  

            A                                    9.8 m/s²                        

            B                                    25 m/s²                         

            C                                    1.67 m/s²                      

i. Calculate the velocity of object falling freely towards B in 5th second.

→ Solution:

    Value of 'g' at surface of B = 25 m/s² 

    initial velocity = u = 0 m/s

    time = t = 5 seconds

    velocity of object falling = v = ?

We know, For freely falling object

    velocity of object, v = u + gt 

                                     = 0 + 25 × 5 = 125 m/s

ii. If a man can lift 50 kg mass on the surface of the earth, how much mass can he lift on the surface of C?

→ He can lift 50 × ( 9.8 / 1.67 ) kg ≈ 293.41 kg on surfac of C. 

iii. Why are there big craters on the surface of heavenly body of C ?

→ The surface of heavenly bosy of C has craters because it lacks an atmosphere to burn up meteoroids before they hit the surface. When meteoroids, asteroids, or comets collide with heavenly body C, they create craters due to the high-speed impact.

iv. Although the mass of Jupiter is 319 times of mass of the earth, but its acceleration due to gravity does not increase in the same ratio, why?

→ We know that the value of acceleration due to gravity of object on the surface of any heavenly body is given by g = GM / R² , where M is mass and R is radius of heavenly body. This suggests that the value of g is directly proportional to mass but inversely proportional to square of its radius. 

    Although Jupiter's mass is 319 times that of Earth's, its radius is also about 11 times larger. Since R² appears in the denominator, the larger radius significantly reduces the effect of its increased mass on the acceleration due to gravity. As a result, Jupiter's gravity is approximately 2.5 times earth's gravity.

18. In the given condition what will be the gravitational force produceed between P and Q amd betweem Q and R? In which case gravitational force is more by how much? Calculate. [SEE 2080/81 Grade Increment Exam]

SEE 2080 GI

→ See Q. No. 9