A planet having mass equal to that of the earth (M = 6 × 10²⁴ kg) has radius R such that a particle projected from its surface at the speed of light (c = 3 × 10⁸ ms^–1) just fails to escape. Assuming Newton’s Law of gravitation to be valid calculate the radius and mass density of such a planet. Are the numbers realistic?
Note: The radius that you calculated is known as Schwarzschild radius. Actually we need to use theory of general relativity for solving this problem.

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Mass density of the planet is 2 $\times 10^{30}$ kg $m^{- 3}$

Radius of the planet is 8.9 mm

Mass density of the planet is 3 $\times 10^{30}$ $k g m^{- 3}$

Radius of the planet is 9.8 mm

answer is A, C.

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Detailed Solution

$\frac{1}{2} {mc}^{2} - \frac{GMm}{R} = 0 \begin{array}{l} ∴ R = \frac{2 GM}{c^{2}} = \frac{2 \times 6 .67 \times 10^{- 11} \times 6 \times 10^{24}}{{(3 \times 10^{8})}^{2}} \\ = 8 .89 \times 10^{- 3} m = 8 .9 mm (!!) \end{array}$
Mass density $= \frac{6 \times 10^{24}}{\frac{4}{3} \times 3.14 \times {(8.9 \times 10^{- 3})}^{3}} = 2.0 \times 10^{30} {kgm}^{- 3}$
The figures are unrealistic! A body having mass of the earth cannot act like a black hole.