Question:hard

The gravitational field, due to the 'left over part' of a uniform sphere (from which a part as shown, has been 'removed out'), at a very far off point, \(P\), located as shown, would be (nearly):
gravitational field, due to the 'left over part'

Updated On: Apr 24, 2026
  • $\frac{5}{6} \frac{GM}{x^{2}}$
  • $\frac{8}{9} \frac{GM}{x^{2}}$
  • $\frac{7}{8} \frac{GM}{x^{2}}$
  • $\frac{6}{7} \frac{GM}{x^{2}}$
Show Solution

The Correct Option is C

Solution and Explanation

To find the gravitational field due to the 'left over part' of a uniform sphere, from which a smaller part has been removed, we can analyze the situation using the concept of superposition. The complete sphere of mass \(M\) generates a certain gravitational field at point \(P\), and the removed part generates another gravitational field at \(P\). The gravitational field at any point is a vector sum, hence it can be calculated by subtracting the gravitational effect of the removed part from the complete sphere.

  1. First, calculate the gravitational field due to the complete sphere at point \(P\):
    • The gravitational field due to the complete sphere of mass \(M\) is given by: \(E_{\text{complete}} = \frac{GM}{x^2}\).
  2. Next, calculate the gravitational field due to the 'removed part' at point \(P\):
    • Assuming the removed part is also a sphere of radius \(R\) with a uniform mass distribution, its mass \(m\) can be given by the density ratio: \(m = \frac{M}{\frac{4}{3} \pi R^3} \times \frac{4}{3} \pi R^3 = \frac{M}{8}\).
    • The gravitational field due to this mass \(m\) at point \(P\) is: \(E_{\text{removed}} = \frac{Gm}{x^2} = \frac{G \frac{M}{8}}{x^2}\).
  3. Using the principle of superposition, the net gravitational field at \(P\) due to the 'left over part' is:
    • \(E_{\text{left\ over}} = E_{\text{complete}} - E_{\text{removed}} = \frac{GM}{x^2} - \frac{G \frac{M}{8}}{x^2}\).
    • Simplifying, we get: \(E_{\text{left\ over}} = \frac{GM}{x^2} \left( 1 - \frac{1}{8} \right) = \frac{7}{8} \frac{GM}{x^2}\).

Thus, the gravitational field at point \(P\), due to the 'left over part' of the sphere, is \(\frac{7}{8} \frac{GM}{x^2}\), which matches the given correct answer.

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