Question

The energy required to move a satellite of mass ‘m’ from an orbit of radius 2R to an orbit of radius 3R is:
1. mgR
2. (mgR)/12
3. (mgR)/8
4. (mgR)/2

Answer 1

Correct Answer - Option 2 : (mgR)/12

Concept:

• The satellite moves around the earth in an orbit with some velocity.
• Acceleration due to gravity (g) of the earth having mass M is given by:

\(g = \;\frac{{GM}}{{{R^2}}}\)

\(Gravitaional\;force\;by\;earth\;on\;satellite\;\left( F \right) = \frac{{G\;M\;m}}{{{r^2}}}\)

 

Where M is the mass of the earth, R is the radius of the earth, G is the acceleration due to gravity, m is the mass of the satellite, V is the velocity of the satellite and r is the radius of the orbit.

Calculations:

Given that r₁ = 2R, r₂ = 3R

The energy of satellite at radius 2R is given by

\(E_1 = \frac{GMm}{2(r{_1})}\)

\(\implies E_1 = \frac{GMm}{2(2R)}\)

\(\implies E_1 = \frac{GMm}{4R}\)

The energy of satellite at radius 3R is given by -

\(\implies E_2 = \frac{GMm}{2(r{_2})}\)

\(\implies E_2 = \frac{GMm}{2(3R)}\)

\(\implies E_2 = \frac{GMm}{6R}\)

The energy required to move a satellite of from orbit of radius 2R to 3R is 

Δ E = E₁ - E₂

\( Δ E = {\frac{GMm}{4R}}-\frac{GMm}{6R} = \frac{GMm}{12R}\).......(1)

As we know that g = GM/R²

Rg= GM/R putiing this in the  (1) we get,

\( Δ E = \frac{mgR}{12}\)

So, the correct option is (mgR)/12.