Question

A uniform sphere has radius ' \(R\) ' and mass ' \(M\) '. The magnitude of gravitational field at distances ' \(r_1\) ' and ' \(r_2\) ' from the centre are ' \(E_1\) ' and ' \(E_2\) ' respectively. The ratio \(E_1 : E_2\) is (\(r_1 > R\) and \(r_2 < R\))

• A. \(\frac{R^2}{r_1^2 r_2}\)
• B. \(\frac{R^3}{r_1 r_2}\)
• C. \(\frac{R^3}{r_1^2 r_2}\)
• D. \(\frac{R^3}{r_1 r_2^2}\)

Hint:

Outside a uniform sphere, the gravitational field follows the inverse square law; inside, it is directly proportional to the distance from the center.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The gravitational field intensity inside and outside a solid uniform sphere follows different distance-dependent laws.
Step 2: Key Formula or Approach:
1) Field outside a sphere (\(r>R\)): \(E = \frac{GM}{r^2}\).
2) Field inside a sphere (\(r<R\)): \(E = \frac{GMr}{R^3}\).
Step 3: Detailed Explanation:
For distance \(r_1\) (outside, since \(r_1>R\)):
\[ E_1 = \frac{GM}{r_1^2} \]
For distance \(r_2\) (inside, since \(r_2<R\)):
\[ E_2 = \frac{GMr_2}{R^3} \]
We need the ratio \(E_1 : E_2\):
\[ \frac{E_1}{E_2} = \frac{GM/r_1^2}{GMr_2/R^3} \]
The constant \(GM\) cancels out:
\[ \frac{E_1}{E_2} = \frac{1}{r_1^2} \times \frac{R^3}{r_2} \]
\[ \frac{E_1}{E_2} = \frac{R^3}{r_1^2 r_2} \]
Step 4: Final Answer:
The ratio is \(\frac{R^3}{r_1^2 r_2}\).