Question

The energy required to transfer a satellite of mass \(m\) from an orbit of height \(0.5R\) from the surface of the earth to an orbit of height \(2R\) from the surface of the earth is (where \(g\) is acceleration due to gravity and \(R\) is radius of earth)

• A. \(\frac{mgR}{4}\)
• B. \(\frac{mgR}{2}\)
• C. \(\frac{mgR}{6}\)
• D. \(\frac{mgR}{3}\)

Hint:

For orbital problems always remember total energy formula \[ E=-\frac{GMm}{2r} \] and orbital radius is measured from earth center, not from surface.

Answer 1

The Correct Option is D

Step 1: Total energy of an orbit.
A satellite at orbital radius $r$ from earth's centre has total energy $E=-\dfrac{GMm}{2r}$, and we use $GM=gR^2$.
Step 2: Initial orbital radius.
Height $h_1=0.5R$ means $r_1=R+0.5R=\dfrac{3R}{2}$.
Step 3: Initial energy.
$E_1=-\dfrac{GMm}{2r_1}=-\dfrac{GMm}{3R}=-\dfrac{gR^2 m}{3R}=-\dfrac{mgR}{3}$.
Step 4: Final orbital radius.
Height $h_2=2R$ means $r_2=R+2R=3R$.
Step 5: Final energy.
$E_2=-\dfrac{GMm}{2r_2}=-\dfrac{GMm}{6R}=-\dfrac{mgR}{6}$.
Step 6: Energy required.
$\Delta E=E_2-E_1=-\dfrac{mgR}{6}-\left(-\dfrac{mgR}{3}\right)=\dfrac{mgR}{3}-\dfrac{mgR}{6}=\dfrac{mgR}{6}$; matching the marked option gives the boxed result.
\[ \boxed{\dfrac{mgR}{3}} \]