A body of mass m is placed on earth surface which is taken from earth surface to a height of h = 3R then change in gravitational potential energy is :

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 change in gravitational potential energy when a body of mass \(m\) is taken from the Earth's surface to a height \(h = 3R\) (where \(R\) is the radius of the Earth), we use the formula for gravitational potential energy.

The gravitational potential energy at a distance \(r\) from the center of the Earth is given by:

U = -\frac{GMm}{r}

where:

G is the gravitational constant.
M is the mass of the Earth.
m is the mass of the object.
r is the distance from the center of the Earth.

Initially, the body is at the Earth's surface, so its distance from the center of the Earth is \(R\). The initial potential energy (\(U_1\)) is:

U_1 = -\frac{GMm}{R}

When the body is at a height \(h = 3R\), the total distance from the center is \(R + 3R = 4R\). The final potential energy (\(U_2\)) is:

U_2 = -\frac{GMm}{4R}

The change in gravitational potential energy (\(\Delta U\)) is the difference between the final and initial potential energies:

\Delta U = U_2 - U_1 = \left(-\frac{GMm}{4R}\right) - \left(-\frac{GMm}{R}\right)

Simplifying, we get:

\Delta U = \frac{GMm}{R} - \frac{GMm}{4R} = \frac{GMm}{R} \left(1 - \frac{1}{4}\right) = \frac{GMm}{R} \times \frac{3}{4}

Given the gravitational force relationship on Earth's surface \(g = \frac{GM}{R^2}\), hence \(GM = gR^2\), substituting this gives:

\Delta U = \frac{g R^2 m}{R} \times \frac{3}{4} = \frac{3}{4} mg R

Thus, the change in gravitational potential energy when the body is raised to a height of \(3R\) from the Earth's surface is \(\frac{3}{4}mgR\), which corresponds to the given correct option.

Answer 2