The escape velocity of a sphere of mass m is given by (G = Universal gravitational constant; M = Mass of the earth and R = Radius of the e earth)

Question

Two identical particles each of mass $ m $ go around a circle of radius $ a $ under the action of their mutual gravitational attraction. The angular speed of each particle will be:

Answer 1

To determine the escape velocity for a sphere on Earth, we need to understand the concept of escape velocity itself. It refers to the minimum speed required for an object to break free from the gravitational attraction of a celestial body, such as the Earth, without any additional propulsion.

The formula for escape velocity $v_e$ can be derived from the energy conservation principle. The kinetic energy (KE) of the object needs to equal the gravitational potential energy (PE) at the surface of the planet. Therefore, we equate the kinetic energy to gravitational potential energy:

\[\frac{1}{2}mv_e^2 = \frac{G M_e m}{R_e}\]

Where:

$m$ is the mass of the object.
$G$ is the universal gravitational constant.
$M_e$ is the mass of the Earth.
$R_e$ is the radius of the Earth.

We can simplify this formula by canceling $m$ from both sides:

\[\frac{1}{2}v_e^2 = \frac{G M_e}{R_e}\]

By multiplying both sides by 2 to solve for $v_e^2$, we get:

\[v_e^2 = \frac{2 G M_e}{R_e}\]

Taking the square root of both sides gives us the escape velocity:

\[v_e = \sqrt{\frac{2 G M_e}{R_e}}\]

Therefore, the escape velocity of the sphere is represented by the expression \[\sqrt{\frac{2 G M_e}{R_e}}\].

Comparing this result with the given options, the correct answer is:

$\sqrt{\frac{2GM_e}{R_e}}$

This calculation assumes that the effect of air resistance is negligible and that the escape is happening close to the surface of the Earth.

Answer 2