A spaceship of mass mm circles a planet of mass M in an orbit of radius R. How much energy is required to transfer the spaceship to a circular orbit of radius 3R?

ΔE = GmM/3R

Explanation:

The absolute potential energy of an object in a planet's field is given as:

E = -GmM/2r

where,

E = Potential Energy

G = Universal Gravitational Constant

m = mass of spaceship

M = Mass of Planet

r = distance from surface of planet

Therefore, for initial state:

E = E₁ and r = R

E₁ = - GmM/2R

and for final state:

E = E₂ and r = 3R

E₂ = - GmM/6R

So, the required energy will be:

ΔE = E₂ - E₁ = - GmM/6R + GmM/2R

ΔE = GmM(- 1/6R + 1/2R)

ΔE = GmM/3R

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft