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Gravitation

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Question : 25
Total: 25
A rocket is fired ‘vertically’ from the surface of Mars with a speed of 2 km s ‌–1 . If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars =6.4×1023 kg; radius of Mars =3395 km; G=6.67×10–11 N m‌2 kg‌–2.
Solution:  
Let m = mass of the rocket, M = mass of the Mars and R = radius of Mars. Let v be the initial velocity of rocket.
Initial K.E.=
1
2
mv2; Initial P.E. =−
GMm
R
Total initial energy =
1
2
mv2−
GMm
R

Since 20% of K.E. is lost, only 80% is left behind to reach the height.
Therefore,
Total initial energy available
=
80
100
×
1
2
mv2−
GMm
R
=0.4mv2−
GMm
R
If the rocket reaches the highest point which is at a height h from the surface of Mars, its K.E. is zero and P.E. =
−GMm
(R+h)

Using principle of conservation of energy, we have
0.4mv2−
GMm
R
=−
GMm
(R+h)

or
GM
R+h
=
GM
R
−0.4v2=
1
R
[GM−0.4Rv2]
or
R+h
R
=
GM
GM−0.4Rv2

or
h
R
=
GM
GM−0.4Rv2
−1 =
0.4Rv2
GM−0.4Rv2
or h=
0.4×(2×103)2×(3.395×106)2
[(6.67×10−11)×(6.4×1023)−0.4×(3.395×106)×(2×103)2]
m
≈495km.
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