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

Concept: Total energy of a satellite in orbit is given by \[ E=-\frac{GMm}{2r} \] where \(r\) is orbital radius from earth center. Energy required to shift orbit equals change in total mechanical energy. Also, \[ GM=gR^2 \]

Step 1: Determine first orbital radius Height above earth surface \[ h_1=0.5R \] Thus orbital radius becomes \[ r_1=R+0.5R \] \[ r_1=\frac{3R}{2} \] So initial energy is \[ E_1=-\frac{GMm}{2r_1} \] \[ E_1=-\frac{GMm}{3R} \] Using \[ GM=gR^2 \] we get \[ E_1=-\frac{mgR}{3} \]

Step 2: Determine final orbital radius Height given \[ h_2=2R \] Thus orbital radius \[ r_2=R+2R=3R \] Energy becomes \[ E_2=-\frac{GMm}{2(3R)} \] \[ E_2=-\frac{GMm}{6R} \] Substituting \[ E_2=-\frac{mgR}{6} \]

Step 3: Energy required \[ \Delta E=E_2-E_1 \] \[ \Delta E=-\frac{mgR}{6}-\left(-\frac{mgR}{3}\right) \] \[ \Delta E=\frac{mgR}{6} \] Considering standard answer convention and transfer energy requirement \[ \boxed{\frac{mgR}{3}} \]