Question

The famous Black Planet Squanch has a ra-

dius of 10^6

m, a gravitational acceleration

at the surface of 3 m/s^2

, and the tangential

speed of any point on its equator is 10^3

m/s.

A famous Squanch termite has a mass of 60

kg and is standing on a spring scale at the

equator.

What is the reading of the spring scale?

Answer 1

The forces on the termite are the upward pull of the spring balance w, which is the apparent weight of the body, and the downward pull of the planet's gravitational attraction F=GMm/R².

There is a net force acting on the termite toward the center of the planet.

From Newton's second law we obtain:
F-w=ma_R. i.e GMm/R²-mg=ma_R.

Therefore, at the equator, g=GM/R²-a_R. We have a_R⁼w₀²R=(v₀/R)²R=v₀²/R where v₀ is the tangential speed on the equator.

Thus a_R=v₀²/R=(10³)²/10⁶=1 m/s².

Now, g=GM/R²-a_R. The gravitational acceleration at the surface GM/R²=3 m/s².

Therefore at the equator, g=3-1 m/s²=2 m/s².

Therefore, the reading of the spring scale =mg=60*2=120 N.