JEE Advanced 2022 Paper 1 · Physics · Gravitation · hard

A small satellite is launched from the surface of a planet of radius R and mass M in a direction tangential to the surface with speed v0=√(1.2 GM/R).…

Numerical answer

A small satellite is launched from the surface of a planet of radius

R

and mass

M

in a direction tangential to the surface with speed

v_0=\sqrt{1.2\,GM/R}

. Ignore air resistance and rotation of the planet. The largest distance of the satellite from the centre of the planet, in units of

R

, is ______ . Give the answer correct to two decimal places.

Show answer and solution

Answer: 1.50

The launch point is the nearest point (perigee) because the velocity is tangential. Conservation of angular momentum:

v_0R=v_a r

. Conservation of energy:

\tfrac12v_0^2-\dfrac{GM}{R}=\tfrac12v_a^2-\dfrac{GM}{r}

. Let

x=r/R

and use

v_0^2=1.2\,GM/R

0.6-1=\dfrac{0.6}{x^2}-\dfrac1x

, i.e.

0.4x^2-x+0.6=0

. Roots are

x=1

(the launch point) and

x=1.5

. Hence

r_{max}=1.50\,R

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