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• You must have noticed the circus artists doing rope walking. 
• During this act, they carry a long bamboo pole in their hands. 
• The reason for this is that the line joining the centre of gravity and centre of equilibrium must fall within the rope for achieving the stable equilibrium. 
• Thus when an artist finds that he is falling towards left, he shifts bamboo pole towards right, so that his centre of gravity stay undisturbed. 
• Thus he can balance himself on the rope.

D) g downwards

Since the Star is rotating, there is a force pushing it outward (centrifugal force) and a force pulling it inwards, (gravitational force).

If gravitational force (F_G) > Centrifugal force (F_c)

Then the object on the equator remains stuck on the star. 

Mass of Star, M = 2.5 (M_Sun)

= 2.5 × 2 × 10³⁰ kg

= 5 × 10³⁰ kg

Radius of Star = 12 km = 1.2 × 10⁴ m

⇒ F_G = \(\frac {GMm}{R^2}\) = \(\frac {6.67 \times 10^{-11} \times 5 \times 10^{30}}{(1.2 \times 10^4)^2}\)

= \(\frac {6.67 \times 5}{1.44}\) × 10¹¹ × m

= 2.316 × 10¹² × m N

F_c = mr ω²

Where, r = radius of Star = 12 km

ω = angular speed = 2 × π × (1.2)

⇒ F_c = m × 1.2 × 10⁴ × (2.4π)²

= 6.82 × 10⁵ × m N

Since F_c < F_G the object is stuck to Star’s surface.

Acceleration due to gravity,

g = GM/R² = {(6.67 x 10^-11) x (5 x 10³⁰)/(12 x 10³)²}m

= 2.316 x 10¹² ms^-2

Centrifugal acceleration,

= Rω² = (12 x 10³) x (3π)² = 1.066 x 10⁶ ms^-2

As g > rω², the body will remains stuck with the surface of the star

(i) Free fall:

Whenever an object moves under the influence of the force of gravity alone, it is said to be falling freely.

(ii) Acceleration due to gravity:

The acceleration produced in a body due to the gravitational force of the earth is called the acceleration due to gravity.

[Note: On the earth’s surface, the value of the acceleration due to gravity is almost uniform. If a body falls from a low altitude, the value of the acceleration due to gravity is almost the same.]

(iii) Escape velocity:

When a body is thrown vertically upward from the surface of the earth, the minimum initial velocity of the body for which the body is able to overcome the downward pull by the earth and can escape the earth forever is called the escape velocity.

(iv) Centripetal force:

In uniform circular motion of a body, the force acting on the body is directed towards the centre of the circle. This force is called centripetal force.

1. Acceleration due to gravity, 

g = 9.8ms^-2 , Height h = 16 × 10³m Radius of the earth, 

R = 6400 × 10³ m From the equation g₁ = g \((\frac {R}{R+h})^2\)

= 9.8 \((\frac {6400 \times 10^3}{6400 \times 10^3 + 16 \times 10^3})\)

= 9.75 ms^-2

2. depth, h = 2.8 × 10³m

From the equation g₁ = g \([1-\frac {h}{R}]\)

= 9.8 \([1-\frac {2.8 \times 10^3}{6400 \times10^3}]\)

= 9.796 ms^-2

Acceleration due to gravity is maximum at the poles.

Yes, with larger mass, the value of gravity will be significant.

The value of acceleration due to gravity, g can be considered as constant inside small spaceship orbiting around the earth, so astronaut feels weightlessness.

If space station orbiting around has a large size, the variation in g matters and astronaut will experience gravitational force, hence he can detect gravity. On moon, due larger size, gravity can be detected.

Yes, if the size of the spaceship is large enough for him to detect the variation in g.

(a) Every bodies in the universe attract each other. This attractive force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

(b) \(F\propto\,\frac{m_1m_2}{d^2}\)

\(F=\,\frac{G. m_1m_2}{d^2}\)

(c) m₁ m₂ – Mass of the bodies

d – Distance between the objects 

F – Gravitational force 

G – Gravitational constant 

(d) Mass increases gravitational force increases Distance increases gravitational force decreases.

1) The universal law of gravitation states that every body in the universe attracts other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. 

2) The direction of the force of attraction is along the line joining the centers of the two bodies. 

3) Let two bodies of masses M and M be separated by a distance of ’d’. Then the force of gravitation between them.

4) F_grav  ∝  GM₁M₂/d²  

 5) F_grav = GM₁M₂/d²  

6) G is a proportionality constant called universal gravitational constant and found by Henry cavendish to be G = 6.67 × 10_-11 Nm² Kg^-2

This is because the earth exerts a force of attraction (called gravity) on the stone and pulls it down.

Following are the two applications of universal law of gravitation: 

a) It is used for determining the masses of the sun, the earth, and the moon. 

b) It is also used for discovering new planets and stars.

Distance, d = 1.5×10⁸ km = 1.5×10¹¹m 

Mass of the sun, m = 2×10³⁰ kg 

Mass of the earth, M = 6×10²⁴ kg 

Force of gravitation, F = G(m.M/d²) 

F = 3.57×10²² N

At the centre of the earth, the value of g is zero. Therefore, the weight of the body at the centre of the earth will be zero.

When the stone held in our hand is released, it falls towards the earth because the earth exerts a force of attraction on the stone pulling it towards itself.

F = G.(m.M/d²) 

Force due to gravity, 

F = 1472N

F = G. (m.M/d²) 

m = 50 kg 

M = 120 kg 

d = 10 m 

G = 6.7 × 10^-11 Nm²/kg² 

Substituting the values, 

F = 4.02 × 10^-9

Weight, W = mg 

Mass, m = W/g = 50/9.8 = 5.102kg

(c) of viscous forces causing the speed of satellite and hence height to gradually decrease.