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

Suppose two planets (spherical in shape) of radii

R

and

2 R

, but mass M and 9M respectively have a centre to centre separation 8R 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 plant is

\frac{a}{7} \frac{GM}{R}

then the value of 'a' is ______
[Given: The two planets are fixed in their position]

Answer: 4

Solution

Once past the point where the two pulls cancel, the satellite is drawn in by the heavier planet, so the least speed is the one that brings it to rest exactly there. That point lies at a distance

x

from the centre of

M

with

\frac{M}{x ^{2}} = \frac{9 M}{( 8 R - x ) ^{2}}

, giving

8 R - x = 3 x

and

x = 2 R

. The potential per unit mass at launch, a distance

R

from

M

and

7 R

from

9 M

, is

V_{i} = - \frac{GM}{R} - \frac{9 GM}{7 R} = - \frac{16 GM}{7 R}

and at the neutral point, a distance

2 R

and

6 R

from the two centres,

V_{f} = - \frac{GM}{2 R} - \frac{9 GM}{6 R} = - \frac{2 GM}{R}

Conservation of energy with zero speed at that point gives

\frac{1}{2} v^{2} = V_{f} - V_{i} = \frac{2 GM}{7 R} \Rightarrow v = \frac{4}{7} \frac{GM}{R}

Comparing with the given form,

a = 4

.

Why the neutral point? Between the two planets the net gravitational pull on the satellite is towards

M

for

x < 2 R

and towards

9 M

for

x > 2 R

. So if the satellite just reaches

x = 2 R

with zero speed, the

9 M

planet carries it the rest of the way. Any smaller launch speed means it falls back to

M

. Hence the minimum-speed condition is "arrives at the neutral point with

v = 0

", not "arrives at the far surface with

v = 0

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Numerical answer

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Gravitation

Solution

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