Question

Velocity of a point on the equator of rotating spherical planet is V. The angular velocity of the planet is such that , the value of g at the equator is half of g at the pole. Determine the escape velocity for a polar particle on the planet as a function of v.

• A. $6$v
• B. $2$v
• C. $4$v
• D. $3$v

Answer 1

The correct option is B $2$v

Velocity of a point on the equator of rotating

Spherical planet is $=V$

The angular velocity of the planet is such that

The value of $g$ at the equator is $\frac{1}{2}$ of $g$ at the pole

Now, determination the escape velocity on the planet as a function of $V$

Solution

Let, $\omega$ be the angular velocity of planet and R its radius

$v=\omega R$ at equator $\to \left(1\right)$

$g=g_0-{\omega }^{2}R{\omega }^{2}R=\frac{g_0}{2}{v}^2=\frac{g_{0}R}{2}$

Substituting from $\left(1\right)$

But we know escape velocity

$v_e=\sqrt{2g_{0}R}=\sqrt{4v^2}{v}_e=2v$