Question

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) $

• A. $ 9.2\times 10^{11} m/s^2 $
• B. $ 8.15\times 10^{11} m/s^2 $
• C. $ 7.32\times 10^{11} m/s^2 $
• D. $ 6.98\times 10^{11} m/s^2 $

Answer 1

Answer: (A)

The acceleration due to gravity
$g = \frac{GM}{R^2}$
$ = \frac{6.67\times 10^{-11}\times 1.98 \times 10^{30}}{12000\times 12000}$
$ = 9.2 \times 10^{11}\,m/s^2$