An astronaut takes a ball of mass m from earth to space. He throws the ball into a circular orbit about earth at an altitude of 318.5 km. From earth's surface to the orbit, the change in total mechanical energy of the ball is x (GMₑ m)/(21 Rₑ). The value of x is (take Rₑ=6370 km) :

Solution

Physics sir · 4 steps

Write the altitude as a fraction of Rₑ first. Then it is orbit energy minus surface energy.

1. Orbit radius. The altitude is h = 318.5 km = 6370/20 km = Rₑ/20. So the orbit radius is r = Rₑ + Rₑ/20 = 21Rₑ/20.
2. Energy at surface. On the surface the ball is at rest, so it has only potential energy: E₁ = -(GMₑ m)/Rₑ = -(21 GMₑ m)/21Rₑ.
3. Energy in orbit. In a circular orbit the total energy is half the potential energy: E₂ = -(GMₑ m)/2r = -(20 GMₑ m)/42Rₑ = -(10 GMₑ m)/21Rₑ.

   Δ E = E₂ - E₁ = -(10 GMₑ m)/21Rₑ + (21 GMₑ m)/21Rₑ = (11 GMₑ m)/21Rₑ

4. So. Matching with x(GMₑ m)/21Rₑ gives x = 11, option (b).

The idea to keep. A body at rest on the surface has E = -GMm/R, and a body in circular orbit has E = -GMm/2r. The energy you must supply is the difference between the two.

Answer from the NTA final answer key. Solution by PrepWiser.