AP Physics B & C – Multiple Choice Questions (for practice) on Gravitation

The essential points you have to remember in respect of gravitation were discussed in the post dated 9^th May 2008. As promised in the post, we will discuss some multiple choice questions involving gravitation.

(1) A simple pendulum has a time period T when on the earth’s surface. The time period of the same pendulum when it is inside an artificial satellite orbiting around the earth at an altitude equal to the radius of the earth will be

(a) T/4

(b) T/2

(c) 2T

(d) 4T

(e) infinite

In an artificial satellite orbiting the earth, bodies will be weightless. Therefore, there is no restoring force on the bob to oscillate the pendulum. Since the pendulum will not oscillate, its period is infinite [Option (e)].

[To put this in a different manner, the effective value of acceleration due to gravity ‘g’ inside the satellite is zero. On substituting this in the expression for the period, T = 2π√( ℓ/g), you obtain it as infinite].

(2) Infinite number of identical spheres of mass 1 kg each are placed along the X-axis with their centres at x = 1 m, 2 m, 4 m, 8 m, 16 m,…… The magnitude of the resultant gravitational field due to these masses at the origin in terms of the gravitational constant G is

(a) 3G/4

(b) 4G/3

(c) G/2

(d) G/4

(e) infinite

The magnitude of the resultant gravitational field at the origin is given by the sumof the forces on unit mass placed at the origin. These forces being in the same direction, the magnitude of the reultant field (F) is given by

F = G×1/1² + G×1/2² + G×1/4² + G×1/8² +……..

Thus F = G(1 + 1/4 + 1/16 + 1/64 +……..) = 4G/3.

(3) The velocity of escape from the earth’s surface is nearly 11.2 kms^–1. If a body is projected at an angle of 60º with the vertical, its velocity of escape will be

(a) 11.2 kms^–1

(b) 11.2×(2/3) kms^–1

(c) 11.2×(√3/2) kms^–1

(d) 11.2×(2/√3) kms^–1

(e) 11.2/2 kms^–1

The escape velocity is the minimum velocity of projection of a body so as to make it escape into outer space. A body in the gravitational field of the earth has negative gravitational potential energy (equal to –GM/r, with usual notations). By supplying kinetic energy (which is always positive), the total energy is to be made equal to zero to make the body free from the gravitational pull and escape into outer space. So, it is the kinetic energy and hence the magnitude of the velocity of projection that matters and not the direction of projection. The answer therefore is 11.2 kms^–1.

(4) If the radius of the earth were to decrease by 0.1 % without any change in its mass, the acceleration due to gravity on the earth’s surface would

(a) increase by 0.1 %

(b) decrease by 0.1 %

(c) increase by 0.2 %

(d) decrease by 0.2 %

(e) increase by 0.05 %

The expression for the surface value of acceleration due to gravity (g) is

g = GM/R² where where G is the gravitational constant, M is the mass of the earth and R is its radius.

The fractional change in g on changing the quantities on the right hand side is given by δg /g = δG /G + δM /M – 2δR /R

Since G and M have constant values, the first two terms on the right hand side are zero so that δg /g = – 2 δR /R

Therefore, percentage change in g = – 2×(percentage change in R).

The percentage change in the radius R is – 0.1 %. [Negative sign since the change is decrement]

Therefore, percentage change in g = – 2×(– 0.1) = 0.2 %. Since the sign is positive, this is an increment and the correct option is (c).

(5) Imagine a body at rest at height R from the earth’s surface, where R is the radius of the earth. If it falls freely under the gravitational pull of the earth, what will be its velocity just before it hits the earth’s surface where the acceleration due to gravity is g? Neglect the air resistance for the sake of simplicity of the problem.[In a real situation you cannot neglect the air resistance which may even burn the entire body before it reaches the ground!]

(a) √(gR)

(b) √(gR)

(c) √(gR/2)

(d) √(gR/3)

(e) √(2gR/3)

If the height ‘h’ is negligible compared to the radius of the earth, the velocity on hitting the ground would be √(2gh) which you obtain by writing ½ mv² = mgh . But you cannot replace h with R since the value of acceleration due to gravity changes appreciably over the path of the body.

When the body is at altitude R, its distance from the centre of the earth is 2R and its gravitational potential energy is – GMm/2R where m is the mass of the body. On falling down, its kinetic energy increases and its gravitational potential energy decreases (larger negative value).

At the surface of the earth, its gravitational potential energy is – GMm/R.

Therefore, change in gravitational potential energy = –GMm/R – (–GMm/2R) = – GMm/2R. The negative sign shows that the change is a decrement.

The corresponding increment in the kinetic energy is ½ mv² where ‘v’ is the velocity on hitting the ground.

Therefore we have GMm/2R = ½ mv² from which v = √(GM/R) = √(gR) since g = GM/R²

(6) Two planets have radii R₁ and R₂ and mean densities d₁ and d₂ respectively. The ratio of the accelerations due to gravity on their surfaces is

(a) R₁²d₁ : R₂²d₂

(b) R₁d₁ : R₂d₂

(c) R₁d₂ : R₂d₁

(d) R₁d₂² : R₂d₁²

(e) R₁³d₁ : R₂³d₂

The surface value of acceleration due to gravity (g) is given by

g = GM /R²

Since M = (4/3) πR³d where d is the mean density of the planet, we have

g = G×(4/3) πR³d /R² = G×(4/3) πRd

Therefore, g is directly proportional to the product Rd so that the required ratio is R₁d₁ : R₂d₂ [Option (b)].

We will discuss more questions on gravitation in due course.

We have to do the best we can. This is our sacred
human responsibility. 

– Albert Einstein