Suppose two planets (spherical in shape) in radii R and 2R, but mass M and 9M respectively have a centre to centre separation 8 R as shown in the figure. A satellite of mass 'm' is projected from the surface of the planet of mass 'M' directly towards the centre of the second planet. The minimum speed 'v' required for the satellite to reach the surface of the second planet is \sqrt a \over 7GM \over R then the value of 'a' is \underline\hspace2cm . [Given : The two planets are fixed in their position]

Gravitation PYQ — JEE Main 2021 · 27 Jul · Shift 1 · Q67 with Solution

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Correct answer: 4

Set up the geometry

Let the smaller planet have:

mass $M$
radius $R$

The larger planet has:

mass $9M$
radius $2R$

The distance between their centres is $8R$ .

A satellite is launched from the surface of the first planet toward the second planet along the line joining the centres.

So if $x$ is the distance of the satellite from the centre of the first planet, then:

start point: $x=R$
surface of second planet: distance from second centre $=2R$

Since second centre is at $x=8R$ , its near surface is at $x=8R-2R=6R.$

Thus the satellite moves from $x=R$ to $x=6R$ .

Write the gravitational potential energy

At any point $x$ between the planets, gravitational potential energy of the satellite is $U(x)=-\frac{GMm}{x}-\frac{G(9M)m}{8R-x}.$

So, $U(x)=-GMm\left(\frac{1}{x}+\frac{9}{8R-x}\right).$

Find the point of maximum potential energy

For minimum launch speed, the satellite must just be able to cross the point where potential energy is maximum.

So differentiate $U(x)$ : $\frac{dU}{dx}=-GMm\left(-\frac{1}{x^2}+\frac{9}{(8R-x)^2}\right).$

Set $\frac{dU}{dx}=0$ : $-\frac{1}{x^2}+\frac{9}{(8R-x)^2}=0$ $\frac{9}{(8R-x)^2}=\frac{1}{x^2}$ $\frac{3}{8R-x}=\frac{1}{x}$ $3x=8R-x$ $4x=8R$ $x=2R.$

So the potential energy is maximum at $x=2R$ .

Compute potential energy at launch point and barrier point

At launch point $x=R$

=-\frac{GMm}{R}\left(1+\frac{9}{7}\right) =-\frac{16}{7}\frac{GMm}{R}.$$ ### At barrier point $x=2R$ $$U(2R)=-GMm\left(\frac{1}{2R}+\frac{9}{6R}\right) =-\frac{GMm}{R}\left(\frac12+\frac32\right) =-2\frac{GMm}{R}.$$ --- 5. **Apply energy conservation for minimum speed** For minimum speed, the satellite just reaches the maximum-potential point with zero speed there. So, $$\frac12 mv^2 + U(R) = U(2R).$$ Hence, $$\frac12 mv^2 = U(2R)-U(R).$$ Substitute values: $$\frac12 mv^2 = -2\frac{GMm}{R} - \left(-\frac{16}{7}\frac{GMm}{R}\right)$$ $$\frac12 mv^2 = \left(\frac{16}{7}-2\right)\frac{GMm}{R}$$ $$\frac12 mv^2 = \frac{2}{7}\frac{GMm}{R}.$$ Therefore, $$v^2=\frac{4}{7}\frac{GM}{R}$$ $$v=\sqrt{\frac{4}{7}\frac{GM}{R}}.$$ This is of the form $$v=\sqrt{\frac{a}{7}\frac{GM}{R}}.$$ So, $$a=4.$$ --- 6. **Comparison with stored answer** Derived answer: $4$. Stored correct answer: $4$. They match.

More from Gravitation

View written solutionFree

Correct answer: 4

Set up the geometry

Let the smaller planet have:

mass $M$
radius $R$

The larger planet has:

mass $9M$
radius $2R$

The distance between their centres is $8R$ .

A satellite is launched from the surface of the first planet toward the second planet along the line joining the centres.

So if $x$ is the distance of the satellite from the centre of the first planet, then:

start point: $x=R$
surface of second planet: distance from second centre $=2R$

Since second centre is at $x=8R$ , its near surface is at $x=8R-2R=6R.$

Thus the satellite moves from $x=R$ to $x=6R$ .

Write the gravitational potential energy

At any point $x$ between the planets, gravitational potential energy of the satellite is $U(x)=-\frac{GMm}{x}-\frac{G(9M)m}{8R-x}.$

So, $U(x)=-GMm\left(\frac{1}{x}+\frac{9}{8R-x}\right).$

Find the point of maximum potential energy

For minimum launch speed, the satellite must just be able to cross the point where potential energy is maximum.

So differentiate $U(x)$ : $\frac{dU}{dx}=-GMm\left(-\frac{1}{x^2}+\frac{9}{(8R-x)^2}\right).$

Set $\frac{dU}{dx}=0$ : $-\frac{1}{x^2}+\frac{9}{(8R-x)^2}=0$ $\frac{9}{(8R-x)^2}=\frac{1}{x^2}$ $\frac{3}{8R-x}=\frac{1}{x}$ $3x=8R-x$ $4x=8R$ $x=2R.$

So the potential energy is maximum at $x=2R$ .

Compute potential energy at launch point and barrier point

At launch point $x=R$

=-\frac{GMm}{R}\left(1+\frac{9}{7}\right) =-\frac{16}{7}\frac{GMm}{R}.$$ ### At barrier point $x=2R$ $$U(2R)=-GMm\left(\frac{1}{2R}+\frac{9}{6R}\right) =-\frac{GMm}{R}\left(\frac12+\frac32\right) =-2\frac{GMm}{R}.$$ --- 5. **Apply energy conservation for minimum speed** For minimum speed, the satellite just reaches the maximum-potential point with zero speed there. So, $$\frac12 mv^2 + U(R) = U(2R).$$ Hence, $$\frac12 mv^2 = U(2R)-U(R).$$ Substitute values: $$\frac12 mv^2 = -2\frac{GMm}{R} - \left(-\frac{16}{7}\frac{GMm}{R}\right)$$ $$\frac12 mv^2 = \left(\frac{16}{7}-2\right)\frac{GMm}{R}$$ $$\frac12 mv^2 = \frac{2}{7}\frac{GMm}{R}.$$ Therefore, $$v^2=\frac{4}{7}\frac{GM}{R}$$ $$v=\sqrt{\frac{4}{7}\frac{GM}{R}}.$$ This is of the form $$v=\sqrt{\frac{a}{7}\frac{GM}{R}}.$$ So, $$a=4.$$ --- 6. **Comparison with stored answer** Derived answer: $4$. Stored correct answer: $4$. They match.

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