Mass of the earth is M and its radius is R. An object of mass m is placed on the surface of earth. Find the work done in lifting the object through a height \(\frac{R}{2}\) above the surface of earth.
Work Done Lifting Object Above Earth's Surface
This question asks us to calculate the work done to lift an object of mass m from the surface of the Earth to a certain height above the surface. The Earth has mass M and radius R. The height is given as \( \frac{R}{2} \) above the surface.
Understanding Work Done Against Gravity
When we lift an object against gravity, we are doing work to increase its gravitational potential energy. The work done by an external force in moving an object from point A to point B against a conservative force like gravity is equal to the change in the potential energy of the object between these two points. The change in potential energy is given by \( \Delta U = U_{final} - U_{initial} \).
The gravitational potential energy \(U\) of an object of mass \(m\) at a distance \(r\) from the center of the Earth (mass \(M\)) is given by the formula:
\[ U = -\frac{GMm}{r} \]
where \(G\) is the universal gravitational constant.
Calculating Initial and Final Potential Energy
Initial Position: The object is initially placed on the surface of the Earth. The distance from the center of the Earth to the surface is equal to the Earth's radius, \(R\). So, the initial distance is \(r_{initial} = R\).
The initial gravitational potential energy \(U_{initial}\) is:
\[ U_{initial} = -\frac{GMm}{R} \]
Final Position: The object is lifted to a height \( \frac{R}{2} \) above the surface of the Earth. The distance from the center of the Earth to this final position is the Earth's radius plus the height.
So, the final distance is \(r_{final} = R + \frac{R}{2} = \frac{2R}{2} + \frac{R}{2} = \frac{3R}{2}\).
The final gravitational potential energy \(U_{final}\) is:
\[ U_{final} = -\frac{GMm}{r_{final}} = -\frac{GMm}{\frac{3R}{2}} = -\frac{2GMm}{3R} \]
Calculating the Work Done
The work done \(W\) in lifting the object from the initial position (surface) to the final position (height \(\frac{R}{2}\) above surface) is the change in potential energy:
\[ W = U_{final} - U_{initial} \]
Substitute the values of \(U_{final}\) and \(U_{initial}\) we calculated:
\[ W = \left(-\frac{2GMm}{3R}\right) - \left(-\frac{GMm}{R}\right) \]
\[ W = -\frac{2GMm}{3R} + \frac{GMm}{R} \]
To combine these terms, find a common denominator, which is \(3R\):
\[ W = -\frac{2GMm}{3R} + \frac{3 \times GMm}{3 \times R} \]
\[ W = -\frac{2GMm}{3R} + \frac{3GMm}{3R} \]
\[ W = \frac{-2GMm + 3GMm}{3R} \]
\[ W = \frac{(3-2)GMm}{3R} \]
\[ W = \frac{1 \times GMm}{3R} \]
\[ W = \frac{GMm}{3R} \]
Therefore, the work done in lifting the object through a height \( \frac{R}{2} \) above the surface of Earth is \( \frac{GMm}{3R} \).
Summary of Calculation
| Quantity | Formula/Value |
|---|---|
| Mass of Earth | \(M\) |
| Radius of Earth | \(R\) |
| Mass of object | \(m\) |
| Initial distance from center | \(r_{initial} = R\) |
| Final distance from center | \(r_{final} = R + \frac{R}{2} = \frac{3R}{2}\) |
| Initial Potential Energy | \(U_{initial} = -\frac{GMm}{R}\) |
| Final Potential Energy | \(U_{final} = -\frac{GMm}{3R/2} = -\frac{2GMm}{3R}\) |
| Work Done (Change in Potential Energy) | \(W = U_{final} - U_{initial}\) |
| Calculated Work Done | \(W = \frac{GMm}{3R}\) |
This result matches one of the given options.
Revision Table: Gravitational Concepts
| Concept | Description | Formula |
|---|---|---|
| Gravitational Force (between two masses \(M_1\), \(M_2\) at distance \(r\)) | Attractive force between any two objects with mass. | \( F = \frac{GM_1M_2}{r^2} \) |
| Gravitational Potential Energy (of mass \(m\) at distance \(r\) from mass \(M\)) | Energy stored in the gravitational field due to the object's position. It's negative and approaches zero as \(r \to \infty\). | \( U = -\frac{GMm}{r} \) |
| Work Done by External Force | Work required to move an object against a force; equals change in potential energy for conservative forces. | \( W = \Delta U = U_{final} - U_{initial} \) |
Additional Information: Gravitational Potential Energy and Work Done
The formula for gravitational potential energy \( U = -\frac{GMm}{r} \) is defined with the reference point of zero potential energy being at infinite distance (\(r \to \infty\)). As an object gets closer to a massive body, its potential energy becomes more negative. Lifting an object means moving it further away from the Earth, increasing its distance \(r\), which makes the potential energy less negative, thus increasing it (\(U_{final} > U_{initial}\) for positive work). The positive value of the work done calculation confirms that external work is needed to move the object to a higher potential energy state.
The formula \(U = mgh\) is often used for calculating potential energy change near the Earth's surface, where \(g\) is the acceleration due to gravity (\(g = \frac{GM}{R^2}\)) and \(h\) is the height above the surface. However, this formula is an approximation valid only for heights \(h\) much smaller than the radius of the Earth \(R\). Since the height in this problem (\(\frac{R}{2}\)) is comparable to the radius, we must use the more general formula for gravitational potential energy based on the distance from the Earth's center.
Important Questions from Gravitational potential energy
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(i.e. Vp - Vsurface)
Which term is used for celestial bodies that revolve around the sun in highly elliptical orbit ?
