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]
[If, g = 10 ms–2 and radius of earth = 6400 km]
- 48 MJ
- 24 MJ
- 36 MJ
- 12 MJ
The Correct Option is A
Approach Solution - 1
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.
Approach Solution -2
The gain in potential energy of the object,
ΔU=Uf–Ui
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
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Concepts Used:
Center of Mass
The center of mass of a body or system of a particle is defined as a point where the whole of the mass of the body or all the masses of a set of particles appeared to be concentrated.
The formula for the Centre of Mass:
Center of Gravity
The imaginary point through which on an object or a system, the force of Gravity is acted upon is known as the Centre of Gravity of that system. Usually, it is assumed while doing mechanical problems that the gravitational field is uniform which means that the Centre of Gravity and the Centre of Mass is at the same position.