Calculate the acceleration due to gravity on the surface of a pulsar of mass $ M = 1.98 \times 10^{30}\, kg $ and radius $ R = 12 \,km $ rotating with time period $ T = 0.041 \,s. (G = 6.67 \times 10^{-11}MKS) $
- $ 9.2\times 10^{11} m/s^2 $
- $ 8.15\times 10^{11} m/s^2 $
- $ 7.32\times 10^{11} m/s^2 $
- $ 6.98\times 10^{11} m/s^2 $
The Correct Option is A
Solution and Explanation
$g = \frac{GM}{R^2}$
$ = \frac{6.67\times 10^{-11}\times 1.98 \times 10^{30}}{12000\times 12000}$
$ = 9.2 \times 10^{11}\,m/s^2$
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Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].