Two planets have masses M and 16 M and their radii are a and 2 a , respectively. The separation between the centres of the planets is 10 a . A body of mass m is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firing speed needed is :

Gravitation PYQ — JEE Main 2020 · 6 Sep · Shift 2 · Q54 with Solution

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Correct answer: C

Given data

Smaller planet: mass $M$ , radius $a$
Larger planet: mass $16M$ , radius $2a$
Distance between centres $=10a$
A body of mass $m$ is projected from the surface of the larger planet toward the smaller one.

We need the minimum speed so that the body just reaches the surface of the smaller planet.

Idea of minimum speed

For the body to go from one planet to the other, it must cross the point along the line joining centres where the net gravitational force is zero. That point corresponds to the maximum of potential energy along the path.

So the minimum launch speed is obtained by making the body just reach this highest potential point with zero speed.

Thus,

\frac12 m v_{\min}^2 = U_{\max}-U_{\text{start}}

where $U$ is the gravitational potential energy of the body due to both planets.

Locate the neutral point

Let the neutral point be at distance $x$ from the centre of the larger planet. Then its distance from the smaller planet is $10a-x$ .

At neutral point,

\frac{G(16M)}{x^2}=\frac{GM}{(10a-x)^2}

Cancelling $GM$ ,

\frac{16}{x^2}=\frac{1}{(10a-x)^2}

Taking square root,

\frac{4}{x}=\frac{1}{10a-x}

So,

4(10a-x)=x

40a-4x=x

5x=40a Rightarrow x=8a

Hence the neutral point is:

$8a$ from larger planet centre
$2a$ from smaller planet centre

Potential energy at start point

The body starts from the surface of the larger planet on the side facing the smaller one.

So its distances from the centres are:

from larger planet: $2a$
from smaller planet: $10a-2a=8a$

Therefore,

U_{\text{start}}=-\frac{G(16M)m}{2a}-\frac{GMm}{8a}

U_{\text{start}}=-\frac{8GMm}{a}-\frac{GMm}{8a}

U_{\text{start}}=-\frac{65}{8}\frac{GMm}{a}

Potential energy at the neutral point

At the neutral point, distances are:

from larger planet: $8a$
from smaller planet: $2a$

Thus,

U_{\max}=-\frac{G(16M)m}{8a}-\frac{GMm}{2a}

U_{\max}=-\frac{2GMm}{a}-\frac{GMm}{2a}

U_{\max}=-\frac{5}{2}\frac{GMm}{a}

Minimum kinetic energy needed

\frac12 m v_{\min}^2=U_{\max}-U_{\text{start}}

=\left(-\frac{5}{2}\frac{GMm}{a}\right)-\left(-\frac{65}{8}\frac{GMm}{a}\right)

=\left(-\frac{20}{8}+\frac{65}{8}\right)\frac{GMm}{a}

=\frac{45}{8}\frac{GMm}{a}

So,

\frac12 m v_{\min}^2=\frac{45}{8}\frac{GMm}{a}

v_{\min}^2=\frac{45}{4}\frac{GM}{a}

v_{\min}=\frac{3}{2}\sqrt{\frac{5GM}{a}}

Match with options

This is Option C.

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

More from Gravitation

View written solutionFree

Correct answer: C

Given data

Smaller planet: mass $M$ , radius $a$
Larger planet: mass $16M$ , radius $2a$
Distance between centres $=10a$
A body of mass $m$ is projected from the surface of the larger planet toward the smaller one.

We need the minimum speed so that the body just reaches the surface of the smaller planet.

Idea of minimum speed

So the minimum launch speed is obtained by making the body just reach this highest potential point with zero speed.

Thus,

\frac12 m v_{\min}^2 = U_{\max}-U_{\text{start}}

where $U$ is the gravitational potential energy of the body due to both planets.

Locate the neutral point

Let the neutral point be at distance $x$ from the centre of the larger planet. Then its distance from the smaller planet is $10a-x$ .

At neutral point,

\frac{G(16M)}{x^2}=\frac{GM}{(10a-x)^2}

Cancelling $GM$ ,

\frac{16}{x^2}=\frac{1}{(10a-x)^2}

Taking square root,

\frac{4}{x}=\frac{1}{10a-x}

So,

4(10a-x)=x

40a-4x=x

5x=40a Rightarrow x=8a

Hence the neutral point is:

$8a$ from larger planet centre
$2a$ from smaller planet centre

Potential energy at start point

The body starts from the surface of the larger planet on the side facing the smaller one.

So its distances from the centres are:

from larger planet: $2a$
from smaller planet: $10a-2a=8a$

Therefore,

U_{\text{start}}=-\frac{G(16M)m}{2a}-\frac{GMm}{8a}

U_{\text{start}}=-\frac{8GMm}{a}-\frac{GMm}{8a}

U_{\text{start}}=-\frac{65}{8}\frac{GMm}{a}

Potential energy at the neutral point

At the neutral point, distances are:

from larger planet: $8a$
from smaller planet: $2a$

Thus,

U_{\max}=-\frac{G(16M)m}{8a}-\frac{GMm}{2a}

U_{\max}=-\frac{2GMm}{a}-\frac{GMm}{2a}

U_{\max}=-\frac{5}{2}\frac{GMm}{a}

Minimum kinetic energy needed

\frac12 m v_{\min}^2=U_{\max}-U_{\text{start}}

=\left(-\frac{5}{2}\frac{GMm}{a}\right)-\left(-\frac{65}{8}\frac{GMm}{a}\right)

=\left(-\frac{20}{8}+\frac{65}{8}\right)\frac{GMm}{a}

=\frac{45}{8}\frac{GMm}{a}

So,

\frac12 m v_{\min}^2=\frac{45}{8}\frac{GMm}{a}

v_{\min}^2=\frac{45}{4}\frac{GM}{a}

v_{\min}=\frac{3}{2}\sqrt{\frac{5GM}{a}}

Match with options

This is Option C.

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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