The mass density of a planet of radius $ \mathrm{R} $ varies with the distance $ \mathrm{r} $ from its centre as $ \rho(r)=\rho_{0}\left(1-\frac{r^{2}}{R^{2}}\right) $ Then the gravitational field is maximum at:

Question

The mass density of a planet of radius $ \mathrm{R} $ varies with the distance $ \mathrm{r} $ from its centre as $ \rho(r)=\rho_{0}\left(1-\frac{r^{2}}{R^{2}}\right) $ Then the gravitational field is maximum at:, $ r=\sqrt{\frac{3}{4}} R $, $ r=R $, $ r=\frac{1}{\sqrt{3}} R $, $ r=\sqrt{\frac{5}{9}} R $

MARKS App · Accepted Answer

dm = ρ × 4 πx 2 dx = ρ 0 1 - x 2 r 2 × 4 πx 2 dx m = 4 πρ ∫ 0 r x 2 - x 4 R 2 dx s m = 4 πρ 0 r 3 3 - r 5 5 R 2 E = Gm r 2 = G r 2 × 4 πρ 0 r 3 3 - r 5 5 R 2 E = 4 πGρ 0 r 3 - r 3 5 R 2 E is maximum when dE dt = 0 ⇒ dE dr = 4 πGρ 0 1 3 - 3 r 2 5 R 2 = 0 ⇒ r = 5 3 R E max = 4 πGρ 0 × 5 R 3 1 3 - 1 5 × 5 9 E max = 8 5 27 πGρ 0 R

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