A 280kg rocket moving radially outward from the earth\'s surface has a velocity

Question

A 280kg rocket moving radially outward from the earth's surface has a velocity of 1500m/s at an altitude of 4.00e5 m. The mass of the earth is 5.98x10^24 kg and the radius of the earth is 6.37x10^6 m.

a. What is the rocket's orbital velocity at that point?
b. What is the rocket's period at that point?
c. What would be the rocket's escape velocity at that point?
d. What is the rocket's mechanical energy?
e. What is the maximum height above the surface reached by the rocket?

The Answers:
a. 7675.7
b. 5541.8
c. 10855.1
d. -1.618x10^10
e. 532500

I know the answers but i don't know the how you get those answers
Tell me what the units gonna be for the each answer too
Thanks!!

Explanation / Answer

a) orbital velocity = (GM/r) = (GMR^2r/R^2) = (gR^2/r) = R(g/r) = 6.37*10^6*(9.8/6.6*10^6) = 7762 m/s

b) Time period = 2*3.142*6.77*10^6/7762 = 5480.8s

c) escape velocity = 2(orbital sped) = 1.4142*7762 = 10977.02m/s

d) energy = -GMm/r + 0.5mv^2 = 0.5*280*1500^2 - 9.8*250*(6.37*10^6)^2/(6.77*10^6) = -1.43*10^10

e) Change in potential energy = GMm/r - GMm/(r+h) = 0.5*280*1500^2

Height = r+h-R = 5.32*10^5

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