. A spaceship of mass m has its engines switched off and is moving in a circular orbit at height R above the surface of a planet of mass M and radius R. a) Derive an expression for total mechanical energy E of the orbiting spaceship, in terms of G, m, M and R. b) Derive an expression for the minimum speed V the spaceship would need to escape from this orbit into deep space, in terms of system parameters. (The engines can’t fire for the whole trip; they can only give the spaceship one boost so it obtains this velocity. Ignore all other celestial objects.)

Note that GmM/r^2 = ma = MV^2/r

After dividing we have that the velocity of the spaceship will be

V = √GM/R

Now to find the mech. Energy,

But mechanical energy = 1/2 MV^2

Therefore mechanical energy of the spaceship will be

(a) mech. Energy= 1/2m(√GM/R)^2

(b) escape velocity V = √GM/R already calculated above.

the answer is a. at the equator

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.