Question:
When \(g\) is the acceleration due to gravity on earth, the gain in potential energy of an object of mass \(m\) raised from the surface of earth to a height equal to the radius of earth \(R\) is
When \(g\) is the acceleration due to gravity on earth, the gain in potential energy of an object of mass \(m\) raised from the surface of earth to a height equal to the radius of earth \(R\) is
Show Hint
For raising a body from the earth's surface to a height equal to the earth's radius,
\[
\Delta U
=
GMm\left(\frac1R-\frac1{2R}\right)
=
\frac{GMm}{2R}
=
\frac{mgR}{2}.
\]
Do not use \(\Delta U=mgh\) here because \(g\) is not constant over such a large height.
Updated On: Jul 9, 2026
- \(\dfrac{mgR}{2}\)
- \(\dfrac{mgR}{4}\)
- \(mgR\)
- \(2mgR\)
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The Correct Option is A
Solution and Explanation
Concept:
The gravitational potential energy of a mass \(m\) at a distance \(r\) from the centre of the earth is
\[
U=-\frac{GMm}{r}.
\]
Hence, the gain in potential energy is
\[
\Delta U=U_f-U_i.
\]
Step 1: Write the initial and final distances from the centre of the earth. Initially, the object is on the surface of the earth. \[ r_i=R. \] It is raised to a height equal to the radius of the earth. \[ h=R. \] Therefore, \[ r_f=R+R=2R. \]
Step 2: Calculate the change in gravitational potential energy. \[ \Delta U = -\frac{GMm}{2R} - \left(-\frac{GMm}{R}\right). \] \[ \Delta U = \frac{GMm}{R} - \frac{GMm}{2R}. \] \[ \Delta U = \frac{GMm}{2R}. \]
Step 3: Express the answer in terms of \(g\). Since \[ g=\frac{GM}{R^2}, \] we have \[ GM=gR^2. \] Substituting, \[ \Delta U = \frac{gR^2m}{2R}. \] \[ \Delta U = \frac{mgR}{2}. \]
Step 4: Write the final answer. \[ \boxed{\Delta U=\frac{mgR}{2}} \] \[ \boxed{\text{Answer = (A)}} \]
Step 1: Write the initial and final distances from the centre of the earth. Initially, the object is on the surface of the earth. \[ r_i=R. \] It is raised to a height equal to the radius of the earth. \[ h=R. \] Therefore, \[ r_f=R+R=2R. \]
Step 2: Calculate the change in gravitational potential energy. \[ \Delta U = -\frac{GMm}{2R} - \left(-\frac{GMm}{R}\right). \] \[ \Delta U = \frac{GMm}{R} - \frac{GMm}{2R}. \] \[ \Delta U = \frac{GMm}{2R}. \]
Step 3: Express the answer in terms of \(g\). Since \[ g=\frac{GM}{R^2}, \] we have \[ GM=gR^2. \] Substituting, \[ \Delta U = \frac{gR^2m}{2R}. \] \[ \Delta U = \frac{mgR}{2}. \]
Step 4: Write the final answer. \[ \boxed{\Delta U=\frac{mgR}{2}} \] \[ \boxed{\text{Answer = (A)}} \]
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