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WHAT ARE THE IMPORTANT QUESTIONS THAT WILL ASK ON THE EXAM, BASED ON CHAPTER 'GRAVITATION'

1. Find the gravitational force exerted by the earth on a rock weighing 100 kgwt. Weight of the earth is 6 ×10²⁴, radius of the earth is 6.4 × 10⁶ m and g = 10 ms^-2.

Given
Weight of the rock = 100 kgwt
Mass of the rock (m) = 100 kg
Mass of the earth (M_e) = 6 ×10²⁴ kg
Radius of the earth (R) = 6.4 × 10⁶ m
Gravitational constant (G) = 6.67 ×10^-11 Nm²/Kg²
To calculate : Force (F) = ?

2. A body of mass 120 kg is taken to the surface of the moon, where acceleration due to gravity is ¹/6^th of that on the surface of the earth. What will be the weight of the body on the surface of the moon?

Given :
Mass of the body (m) = 120 kg
Acceleration due to gravity (g') = ^9.8/6 = 1.63 m/s²
To calculate weight (W) = ?
Formula to be used : W = m ×  g'
Where g' is the acceleration due to gravity on the surface of the moon.
W = 120 × 1.63
W = 196.0 N

3. Calculate the force exerted by the earth on the moon. Given mass of earth, M_e = 6 1024, mass of the moon, m = 7.4 × 10²² kg, the distance of the moon from the earth = 3.84 × 10⁵ km and G = 6.7 × 10^-11 Nm²/Kg².

Given :
Mass of the earth (M_e) = 6 × 10²⁴ kg
Mass of the moon (m) = 7.4 × 10²² km
Distance (r) = 3.84 × 10⁵ km
             = 3.84 × 10⁸ m
Gravitational constant (G) = 6.7 × 10^-11 Nm²/kg²
To calculate : Force (F) = ?

4. What is the force between two spheres having mass 20 kg each separated by a distance of 50 cm? G = 6.7 × 10^-11 SI units.

Given :

Mass (m₁, m₂) = 20 kg

Distance (r) = 50 cm = 0.50 m

Gravitational constant (G) = 6.7 × 10^-11 Nm²/kg²

To calculate : Force (F) = ?

5. What is the force of gravitation acting between a Maruti car (mass : 1000 kg) and the earth?

Mass of the earth (M) = 6 × 10²⁴ kg

Radius of the earth (R) = 6.4 × 10⁶ m

G = 6.67 × 10^-11 Nm²kg^-2

Using;

We have, m1 = 1000 kg

Mass of the earth m2 (= M) = 6 × 1024 kg

                                         r  = R = 6.4 × 106 m

                                         G = 6.67 × 10^-11 Nm²kg^-2

6. What is the gravitational force of attraction between the sun and the earth?

Mass of the earth = 6 × 10²⁴ kg

Mass of the sun = 2 × 10³⁰ kg

Distance between the sun and the earth = 1.5 × 10¹¹ m

G = 6.67 × 10^-11 Nm²kg^-2

We know

7. When a body of mass 1 kg is dropped from the roof of a house, it gets attracted towards the earth with a certain gravitational force. In accordance with the 3^rd law of motion, the stone also attracts the earth with the same gravitational force towards it. But, why do not we see any appreciable motion of the earth towards the stone? Explain by using the specific data.

Let, Mass of the earth (M₁) = 6 × 10²⁴ kg

Mass of the body (M₂) = 1 kg

Radius of the earth , r (= R) = 6.4 × 10⁶ m

Where F is gravitational force between the earth and the body.

Since the earth has a very large radius, the distance between the centres of the earth and the body is taken as the radius of the earth.

This gravitational force is the force acting on the earth due to the body and is equal to the force acting on the body due to the earth. Though the forces are equal, they act in opposite directions on different bodies in accordance with the 3rd law of motion.

Case (i) Acceleration of the body in the downward direction.

a₁ = ^F/1 kg = 9.8 ms^-2

Distance travelled by the body,

S₁ = ¹/2 gt² = ¹/2 × 9.8 × 1 = 4.9 m  

We have taken t = 1 s and the initial velocity , u = 0

Case (ii) Acceleration of the earth in the upward direction

a₂ = ^F/M₁ = ^9.8/_{6 ×10}²⁴

      = 1.63 × 10^-24 ms^-2

Distance travelled by the earth,

S₂ = ¹/₂ a₂ (1)²  = ¹/₂ × 1.63 × 10^-24

      = 8.15 × 10^-25 m

From (i) and  (ii) we notice that the distance travelled by the earth is negligible small as compared to that travelled by the body under the same given conditions. This clearly explains our inability to notice the motion of the earth.

8. The gravitational force between the objects is 190 N. At what distance should the objects be kept so that the force between these objects becomes 380 N?

We know

Since G, m1 and m2 are not changing, we have

F ? ¹/_d²                ---- (i)

Also,   F₁? ¹/ d₁²    ----- (ii)

From (i) and (ii)  ^F1/_F = d²/d₁²

That is, d²/d₁² = ³⁸⁰/₁₉₀ = 2

or d₁ = ^d/?2

Thus, in order to double the gravitational force of attraction, the given objects be now kept at a distance of d/?2, where d is the original distance between them.

9. The acceleration due to the surface of the earth is 9.8 ms^-2. If the radius of the orbit of the spaceship from the centre of the earth is 2R, where R is radius of the earth, find the acceleration due to gravity at the spaceship.

We know,

Where m, M and R stand for mass of a body, mass of the earth and radius of the earth respectively.

But F = mg

mg = ^GmM/R²

or g = ^GM/R²       ---- (i)

When the distance from the centre of the earth is 2R, then the acceleration due to gravity (g') becomes,

g' = ^GM/(2R)² = ¹/4 ^GM/R²     ----- (ii)

From (i) and (ii) we get,

^g'/g = ¹/4

g' = ^g/₄

    = 9.8 ms^-2

     = 2.45 ms^-2

10. Suppose the mass of the earth increases by 20% but the size of the earth remains the same. What will be the weight of the body on the surface of the earth?

We know

But F = weight (W) = mg

mg = ^GMm/R²

or g = ^GM/R^{2          ------- (i)}

When M increases by 20% then the new mass becomes, M + (20/100) M that is., mass becomes 6M/5

g' = ^6GM/5 × ¹/R²            ------  (ii)

From (i) and (ii) we have, g' = ⁶/₅ (^GM/R²)

                                             = ⁶/₅ g

New weight = mg' = W'

                      W'  = m × ⁶/₅ g = ⁶/₅ W = 1.2 W

Thus, the new weight of the body becomes 1.2 times the original weight of the body.

11. A body is taken from the equator to the poles. What will be the percentage change in the weight of the body? The equatorial radius of the earth is 21 km more than the polar radius of the earth. Polar radius, R_p = 6357 km.

We know,

But F = W = mg,

Where W is weight of the body.

mg = ^GMm/R²

or g = ^GM/R²

using this relation, we can write.

g_e = ^GM/R_e²

and g_p = ^GM/R_p², where ge and gp

stand for the value of g at the equator and the pole respectively. Also , R_e and R_p stand for the equatorial and polar radii respectively.

12. A boy on a roof 49 m high drops a stone. One second later he throws a second stone after the first. They both hit the ground at the same time. With what speed did he throw the second stone?

Given:

Distance (S) = 49 m

Initial velocity (u) for the first stone = 0

Acceleration due to gravity (g) = 9.8 m/s²

To calculate : Initial velocity (u) for the 2^nd stone = ?

Formula to be used : S = ut = ¹/2 gt²

Substituting the value of S, u and g to calculate t, we get,

49 = (0 × t) + (¹/2 × 9.8) (t²)

49 = 4.9 × t²

t² = 10

or  t = 3.162 s

Time  taken by second stone to cover the same vertical distance (49 m) = 3.162 - 1.0 = 2.162 s

(since second stone was thrown 1 s after the first stone)

Let u be the initial velocity of the second stone (to be calculated).

Using S = ut + ¹/2 gt²

49 = u × (2.162) + ¹/2 (9.8) (2.162)²

or 

13. A ball is thrown up with a velocity of 0.5 m/s. How high will it go before it begins to fall?

Given:

Initial velocity (u) = 0.5 ^m/s

Acceleration due to gravity (g) = - 9.8 m/s²

(g is taken to be negative since the body is going against the gravity)

Final velocity (v) = 0 (since the body stop finally)

To calculate : Distance (S) = ?

Formula to be used : v² - u² = 2 gS

Substituting the given values,

(0)² - (0.5)² = 2(- 9.8) (S)

- (0.5 × 0.5) = - 2 × 9.8 × S

or

Solving we get, S = 0.012 m