Question

An object of mass 1 kg is taken to a height from the surface of earth which is equal to three times the radius of earth. The gain in potential energy of the object will be
[If, g = 10 ms^–2 and radius of earth = 6400 km]

• A. 48 MJ
• B. 24 MJ
• C. 36 MJ
• D. 12 MJ

Answer 1

The Correct Option is A

To calculate the gain in potential energy when an object is taken to a height above the surface of the Earth, where this height is equal to three times the radius of the Earth, we use the formula for gravitational potential energy in the gravitational field:

The gravitational potential energy \(U\) at a height \(h\) above the Earth's surface is given by:

\(U = -\frac{GMm}{r + h}\) 

where:

• \(G\) is the gravitational constant.
• \(M\) is the mass of Earth.
• \(m\) is the mass of the object.
• \(r\) is the radius of the Earth.
• \(h\) is the height above the Earth's surface.

Instead of \(U = -\frac{GMm}{r}\) for the surface, the change in potential energy is:

\(\Delta U = GMm \left( \frac{1}{r} - \frac{1}{r+h} \right)\)

At the surface of the Earth, the gravitational potential energy is:

\(U_{\text{surface}} = -\frac{GMm}{r}\)

At a height \(h = 3r\) above the surface, it is:

\(U_{\text{height}} = -\frac{GMm}{r + 3r} = -\frac{GMm}{4r}\)

The gain in potential energy, therefore, is:

\(\Delta U = U_{\text{height}} - U_{\text{surface}} = -\frac{GMm}{4r} - \left(-\frac{GMm}{r}\right)\) \(\Delta U = GMm\left(\frac{1}{r} - \frac{1}{4r}\right) = GMm \left(\frac{3}{4r}\right)\)

Now, using the simplified gravitational potential energy change considering \(g = \frac{GM}{r^2}\), we have:

\(\Delta U = mgr \left(\frac{3}{4}\right) = m \cdot g \cdot r \cdot \frac{3}{4}\)

Substitute \(m = 1 \, \text{kg}\), \(g = 10 \, \text{m/s}^2\), and \(r = 6.4 \times 10^6 \, \text{m}\):

\(\Delta U = 1 \times 10 \times 6.4 \times 10^6 \times \frac{3}{4} = 48 \times 10^6 \, \text{J}\)

Therefore, the gain in potential energy is 48 MJ.

Answer 2

The gain in potential energy of the object,
ΔU=U_f–U_i
ΔU=−\(\frac{GMm}{4R}+\frac{GMm}{R}\)
\(ΔU=\frac{3GMm}{4R}\)
\(ΔU=\frac{3}{4}mgR\)
\(ΔU= 48 x 10^6J\)
\(ΔU= 48 MJ\)
So, the correct option is (A): 48 MJ