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A satellite moves in a circular orbit of radius r around the Earth. A short burn reduces its speed to half of the orbital speed without changing its…

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A satellite moves in a circular orbit of radius

r

around the Earth. A short burn reduces its speed to half of the orbital speed without changing its direction. In the new elliptical orbit, the minimum distance of the satellite from the centre of the Earth is

A. $\dfrac{r}{2}$
B. $\dfrac{r}{4}$
C. $\dfrac{r}{7}$
D. $\dfrac{r}{8}$

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Answer: (C) $\dfrac{r}{7}$

With

v_0^2 = GM/r

, the burn point becomes the apogee (velocity perpendicular to the radius, speed below circular). Let the perigee distance be

xr

with speed

v'

. Angular momentum:

r\cdot\tfrac{v_0}{2} = xr\,v'

, so

v' = \dfrac{v_0}{2x}

. Energy:

\tfrac12\cdot\tfrac{v_0^2}{4} - \dfrac{GM}{r} = \tfrac12\dfrac{v_0^2}{4x^2} - \dfrac{GM}{xr}

. Dividing by

v_0^2

\dfrac18 - 1 = \dfrac{1}{8x^2} - \dfrac1x

, which gives

7x^2 - 8x + 1 = 0

, so

x = 1

(the burn point) or

x = \tfrac17

. Minimum distance

= r/7

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