The spacecraft is designed to leave the surface of Mars with the first stage of
ID: 1344767 • Letter: T
Question
The spacecraft is designed to leave the surface of Mars with the first stage of its propulsion system and be put into Martian orbit. Then, the second stage is used to boost the spacecraft from Martian orbit into an interplanetary trajectory and return to Earth.
If the spacecraft is in Martian orbit at an altitude of 400 km, what is the velocity required to escape the gravitational attraction of Mars. Give your answer in km/s.
Note that the velocity direction and magnitude required to actually return to Earth may be different.
Explanation / Answer
We know that
The diamter of mars is (d) =6794km
Then the radius of the mars is (r) =d/2 =3397km =3397*103m
Mass (M) =6.42*1023kg
The universal gravitational constant (G) =6.67*10-11N.m2/kg2
If the spacecraft is in Martian orbit at an altitude of (h) = 400 km =400*103m
Now the total radius (R) =r+h =3397*103m+400*103m =3797*103m
The velocity required to escape the gravitational attraction of Mars is
v = Sqt(2GM/R) =Sqrt(2*6.42*1023kg*6.67*10-11N.m2/kg2/3797*103m)=Sqrt (0.02255*109) =0.4748*104 =4748m/s =4.748km/s
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