Gravitational Field

Practice Questions:

Question 1: moderate

Two solid spherical planets of equal radii R having masses 4M and 9M their centre are separated by a distance 6R. A projectile of mass m is sent from the planet of mass 4 M towards the heavier planet. What is the distance r of the point from the lighter planet where the gravitational force on the projectile is zero ?

1. 1.4 R

2. 1.8 R

3. 1.5 R

4. 2.4 R

View Answer

The point where the gravitational force on the projectile is zero occurs when the gravitational forces from both planets are equal.

\[
\frac{G \cdot 4M \cdot m}{r^2} = \frac{G \cdot 9M \cdot m}{(6R - r)^2}
\]

Simplifying,

\[
\frac{4}{r^2} = \frac{9}{(6R - r)^2}
\]

Taking the square root:

\[
\frac{2}{r} = \frac{3}{6R - r}
\]

Cross-multiplying:

\[
2(6R - r) = 3r
\]

Solving:

\[
12R - 2r = 3r
\]

\[
5r = 12R
\]

\[
r = \frac{12R}{5}
\]

So, the distance from the lighter planet is \( \frac{12R}{5} \).

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Question 2: moderate

Two bodies, each of mass \(M\), are kept fixed with a separation \(2L\). A particle of mass \(m\) is projected from the mid-point of the line joining their centres, perpendicular to the line. The gravitational constant is \(G\). The correct statement(s) is (are) :

(a) The minimum initial velocity of the mass \(m\) to escape the gravitational field of the two bodies is \(4\sqrt{\frac{GM}{L}}\)

(b) The minimum initial velocity of the mass \(m\) to escape the gravitational field of the two bodies is \(2\sqrt{\frac{GM}{L}}\)

(c) The minimum initial velocity of the mass \(m\) to escape the gravitational field of the two bodies is \(\sqrt{\frac{2GM}{L}}\)

(d) The energy of the mass \(m\) remains constant

1. a, b

2. b, d

3. a, c

4. a, d

View Answer

At the midpoint, potential energy is \(U = -\frac{2GMm}{L}\). For escaping to infinity, total mechanical energy must be at least 0:

\(\frac{1}{2}mv^2 - \frac{2GMm}{L} = 0 ⇒ v = 2\sqrt{\frac{GM}{L}}\). Mechanical energy remains conserved as only gravity acts.

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