Question

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

Answer 1

Answer:

here is your answer

Explanation:

Given that,

Mass of satellite =m

Mass of planet =M

Radius =R

Altitude h=2R

Now,

The gravitational potential energy

P.E=r−Gm

Potential energy at altitude =3RGmM

Orbital velocity v0=R+hGmM

Now, the total energy is

  Ef=21mv02−3RGmM

 Ef=213RGmM−3RGMm

 Ef=3RGmM[21−1]

 Ef=6R−GmM

Now, Ei=Ef

Now, the minimum required energy

  K.E=R

Answer 2

Answer:

Kinetic energy = mV²

Explanation:

In circular orbit of a satellite,

Potential energy (P.E) = −2 × Kinetic Energy(K.E)

P.E = −mV²

So,

Just to escape from the gravitational pull, its total mechanical energy should be zero.

Therefore its kinetic energy should be mV².