A spacecraft is approaching Mars after a long trip from the Earth. Its velocity

Question

A spacecraft is approaching Mars after a long trip from the Earth. Its velocity is such that it is traveling along a parabolic trajectory under the influence of the gravitational force from Mars.  The distance of closest approach will be 290 km above the Martian surface. At this point of closest approach the engines will be fired to slow down the spacecraft and place it in a circular orbit 290 km above the surface.

(a) By what percentage must the speed of the spacecraft be reduced to achieve the desired orbit?

Explanation / Answer

escape velocity on the mars
Ve=sqrt(2*GM/R)...

here G=universal gravitational constant=6.67*10^-11 N.m^2/kg^2..
M is mass of mars=6.419*10^23 kg
and R is rdius of mars = 3.396*10^6 m

Ve=25.21*10^3 m/sec...


altitude is 290 km=0.29*10^6 m

orbital radius r=R+h=4.186 m

orbital velocity is Vo=sqrt(G*M/r)=3.198*10^3 m/s...


change in velocity is Ve-Vo=22.012*10^3 m/s

change is [Ve-Vo/Ve]*100=87.314 % reduced

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