Question

Given that g is the acceleration due to gravity at the surface of the Earth, match the following two columns.

	Column I		Column II
A.	Work done is raising a mass m to a height h = R from the surface of the Earth	P.	$\frac{1}{4}mgR$
B.	Kinetic energy of a satellite of mass m at height h = R	Q.	$\frac{1}{2}mgR$
C.	Difference in energies of two satellites each of mass m but one at height $h_2 = 2R$ and another at height $h_1 = R$	R.	$\frac{1}{12}mgR$
D.	Kinetic energy required to raise a particle of mass m to a height h = R if projected vertically from surface of earth.	S.	None

• A. A→S, B→Q, C→S, D→Q
• B. A→Q, B→P, C→R, D→Q
• C. A→Q, B→R C→Q D→S
• D. A→Q, B→S, C→P, D→Q

Answer 1

Answer: (B)

B) A→Q, B→P, C→R, D→Q

Answer 2

The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM}{R^2}$, where $M$ is the mass of the Earth and $R$ is the radius of the Earth. This relation can be written as $GM = gR^2$.

Let's evaluate each item in Column I:

A. Work done in raising a mass m to a height h = R from the surface of the Earth.

The initial position is at the surface ($r_1 = R$). The potential energy is $U_1 = -\frac{GMm}{R}$.
The final position is at a height $h=R$ from the surface, which is a distance $r_2 = R+h = R+R = 2R$ from the center of the Earth. The potential energy is $U_2 = -\frac{GMm}{2R}$.
The work done against gravity is the change in potential energy:
$W = U_2 - U_1 = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$.
Using $GM = gR^2$, we get $W = \frac{gR^2 m}{2R} = \frac{1}{2}mgR$.
So, A matches with Q.

B. Kinetic energy of a satellite of mass m at height h = R.

A satellite in a circular orbit at a height h has an orbital radius $r = R+h$. For h = R, the orbital radius is $r = R+R = 2R$.
For a stable circular orbit, the gravitational force provides the centripetal force:
$\frac{GMm}{r^2} = \frac{mv^2}{r}$, where v is the orbital speed.
The kinetic energy of the satellite is $K = \frac{1}{2}mv^2$. From the orbital equation, $v^2 = \frac{GM}{r}$.
So, $K = \frac{1}{2}m\left(\frac{GM}{r}\right) = \frac{GMm}{2r}$.
For a satellite at height h = R, $r = 2R$.
$K = \frac{GMm}{2(2R)} = \frac{GMm}{4R}$.
Using $GM = gR^2$, we get $K = \frac{gR^2 m}{4R} = \frac{1}{4}mgR$.
So, B matches with P.

C. Difference in energies of two satellites each of mass m but one at height $h_2 = 2R$ and another at height $h_1 = R$.

The total energy of a satellite in a circular orbit at radius r is $E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$.
For the satellite at height $h_1 = R$, the orbital radius is $r_1 = R+h_1 = R+R = 2R$. The energy is $E_1 = -\frac{GMm}{2r_1} = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}$.
For the satellite at height $h_2 = 2R$, the orbital radius is $r_2 = R+h_2 = R+2R = 3R$. The energy is $E_2 = -\frac{GMm}{2r_2} = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}$.
The difference in energies is $E_2 - E_1 = -\frac{GMm}{6R} - (-\frac{GMm}{4R}) = \frac{GMm}{4R} - \frac{GMm}{6R} = GMm \left(\frac{1}{4R} - \frac{1}{6R}\right) = GMm \left(\frac{3-2}{12R}\right) = \frac{GMm}{12R}$.
Using $GM = gR^2$, the difference is $\frac{gR^2 m}{12R} = \frac{1}{12}mgR$.
So, C matches with R.

D. Kinetic energy required to raise a particle of mass m to a height h = R if projected vertically from surface of earth.

This is the minimum kinetic energy that must be given to the particle at the surface of the Earth so that it just reaches the height $h=R$. Let this kinetic energy be $K_{required}$.
Initial position: surface of the Earth ($r_1 = R$). Initial kinetic energy = $K_{required}$. Initial potential energy $U_1 = -\frac{GMm}{R}$. Total initial energy $E_1 = K_{required} + U_1 = K_{required} - \frac{GMm}{R}$.
Final position: height $h=R$ ($r_2 = R+h = 2R$). At the maximum height, the velocity is zero, so the final kinetic energy $K_2 = 0$. The final potential energy $U_2 = -\frac{GMm}{2R}$. Total final energy $E_2 = K_2 + U_2 = 0 - \frac{GMm}{2R} = -\frac{GMm}{2R}$.
By conservation of mechanical energy, $E_1 = E_2$:
$K_{required} - \frac{GMm}{R} = -\frac{GMm}{2R}$.
$K_{required} = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$.
Using $GM = gR^2$, we get $K_{required} = \frac{gR^2 m}{2R} = \frac{1}{2}mgR$.
So, D matches with Q.

Matching the columns: A → Q B → P C → R D → Q

The correct option is B.