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)

$9.2\times 10^{11} m/s^2$

Answer 2

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