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 280 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 280 km above the surface. By what percentage must the speed of the spacecraft be reduced to achieve the desired orbit? % How would the answer to part (a) change if the distance of closest approach and the desired circular orbit altitude were 600 km instead of 280 km? (Note: The energy of the spacecraft - Mars system for a parabolic orbit is E = O.) No change in percentage. A lesser percentage. A greater percentage.

Explanation / Answer

a) First what is the orbital speed at that height

Satellite motion, circular
V = (GM/R)
G = 6.673e-11 Nm²/kg²
M is mass of central body in kg
R is radius of orbit in m

mars mass is 6.42e23 kg
mars radius is 3400 km or 3.4e6 m

V = ((6.673e-11)(6.42e23)/(3.4e6+0.28e6))

then compare that to it's approach velocity, which you don't have enough info to calculate.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.