Gravitation

The gravitional pull on the astronaut is,

F_G  =  ${{GMm} \over {{{\left( {R + h} \right)}^2}}}$

The satellite is moving so fast around the earth that whenever it trying to fall on the earch it is missing the earth and it will keep going arround the earth. On the satellite the astronaut feels weightless because the satellite is constantly freefalling.

If gravity were absent, the satellite would move in a straight line through space instead of orbiting the Earth. In this scenario, the astronaut would still feel weightless, but for a different reason: there would be no gravitational force acting on the satellite in the absence of gravity.

Option B : $ 0 < F_G < \frac{GMm}{R^2} $

The gravitational force $ F_G $ acting on the astronaut while in orbit is less than the gravitational force at the Earth's surface, but it is not zero. The force of gravity decreases with the square of the distance from the center of the Earth. Since the astronaut is at a distance ( R + h ) from the center of the Earth, where ( h ) is the altitude of the orbit, the gravitational force will be less than at the Earth's surface but greater than zero.

So, both Option B and Option D are correct. Option B correctly states that the gravitational force is less than at the Earth's surface but greater than zero, and Option D gives the precise formula for calculating that force.

Here we will use the law of conservation of energy as there are no non-conservative force involved so the total energy of mass m remains constant.

For minimum initial velocity to escape the gravitational field, the velocity at infinity will be zero.

And we know PE at infinity is zero.

By energy conservation,

$P{E_i} + K{E_i} = P{E_f} + K{E_f}$

$ \Rightarrow {{ - GMm} \over L} - {{GMm} \over L} + {1 \over 2}m{v^2} = 0 + {1 \over 2}m{(0)^2}$

$ \Rightarrow - {{2Gm} \over L}M + {1 \over 2}m{v^2} = 0$

$ \Rightarrow {1 \over 2}{V^2} = {{2Gm} \over L}$

$ \Rightarrow V = 2\sqrt {{{GM} \over L}} $

Hence, options (B) and (D) are correct.

Escape velocity, expressed as

$v_e = \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} = \sqrt{\frac{2 \mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R}^3 \rho}{\mathrm{R}}}$

can be simplified to

$\sqrt{\frac{8 \pi \mathrm{G} \rho}{3}} \mathrm{R}.$

The surface area, $A_s$, is given by

$A_s = 4 \pi R^2.$

Considering planet Q's surface area is 4 times that of planet P:

$\mathrm{A}_{\mathrm{Q}} = 4 \mathrm{A} = 4 \pi \mathrm{R}_{\mathrm{Q}}^2$

$\mathrm{~A}_{\mathrm{P}} = \mathrm{A} = 4 \pi \mathrm{R}_{\mathrm{P}}^2$

therefore,

$4 \mathrm{R}_{\mathrm{P}}^2 = \mathrm{R}_{\mathrm{Q}}^2.$

Solving for $R_Q$ :

$\Rightarrow \quad R_{\mathrm{Q}} = 2 \mathrm{R}_{\mathrm{P}} \quad \text{........(i)}.$

For planet R, with mass $\mathrm{M}_{\mathrm{R}} = \mathrm{M}_{\mathrm{P}} + \mathrm{M}_{\mathrm{Q}}$:

$\rho \cdot \frac{4}{3} \pi \mathrm{R}_{\mathrm{R}}^3 = \rho \cdot \frac{4}{3} \pi (\mathrm{R}_{\mathrm{P}}^3 + \mathrm{R}_{\mathrm{Q}}^3)$

Since

$\mathrm{R}_{\mathrm{Q}} = 2\mathrm{R}_{\mathrm{P}},$

we have :

$\mathrm{R}_{\mathrm{R}}^3 = \mathrm{R}_{\mathrm{P}}^3 + 8 \mathrm{R}_{\mathrm{P}}^3 = 9 \mathrm{R}_{\mathrm{P}}^3.$

Thus,

$\mathrm{R}_{\mathrm{R}} = (9)^{1/3} \mathrm{R}_{\mathrm{P}} \quad \text{........(ii)}.$

From (i) and (ii), it follows that :

$ R_R > R_Q > R_P.$

Since

$V_e \propto \mathrm{R},$

it implies that :

$\mathrm{V}_{\mathrm{R}} > \mathrm{V}_{\mathrm{Q}} > \mathrm{V}_{\mathrm{P}}.$

Also,

$\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{V}_{\mathrm{P}}} = \frac{\mathrm{R}_{\mathrm{R}}}{\mathrm{R}_{\mathrm{P}}} = 9^{1/3},$

and

$\frac{\mathrm{V}_{\mathrm{P}}}{\mathrm{V}_{\mathrm{Q}}} = \frac{\mathrm{R}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{Q}}} = \frac{1}{2}.$