How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Radius of a Planet

The cosmonauts who landed at the pole found that the force of gravity there is (0.01) of that on the Earth, while the duration of the day on the planet is the same as that on Earth. It turned out that besides that the force of gravity on the equator is zero. Find the radius (R) of the planet.

Discussion:

For a body of mass (m) resting on the equator of a planet of radius (R), which rotates at an angular velocity (\omega), the equation of motion has the form $$m\omega^2R=mg'-N$$ where (N) is the normal reaction of the planet surface, and (g'=0.001g) is the free-fall acceleration on the planet.
The bodies on the equator are assumed to be weightless i.e. (N=0).
We know, (w=2\frac{\pi}{T}), where (T) is the period of revolution of the planet.
Hence we obtain $$R=\frac{T^2}{4\pi^2}g'$$
Substituting the value of (T=8.610^4s) and (g'=0.1m/s^2), we get $$R=1.810^7 Km$$

# Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy