Gravitation — JEE Main practice

1. The total mechanical energy of a satellite of mass $m$ moving in a circular orbit of radius $r$ is

$$E=K+U,$$

where

$$U=-\frac{GMm}{r}.$$

2. For a circular orbit, the gravitational force provides the centripetal force:

$$\begin{aligned} \frac{mv^2}{r} &=\frac{GMm}{r^2}. \end{aligned}$$

Hence,

$$\begin{aligned} \frac{1}{2}mv^2 &=\frac{GMm}{2r}. \end{aligned}$$

3. Therefore, the total energy is

$$\begin{aligned} E &=\frac{GMm}{2r}-\frac{GMm}{r} \\ &=-\frac{GMm}{2r}. \end{aligned}$$

4. Using

$$g=\frac{GM}{R_E^2},$$

we get

$$GM=gR_E^2.$$

Hence,

$$E = -\frac{mgR_E^2}{2r}.$$

5. The given values are

$$m=100\ \mathrm{kg},\qquad r_1=1.5R_E,\qquad r_2=3R_E.$$

The energy required is

$$\Delta E=E_2-E_1.$$

6. Substituting the energies,

$$\begin{aligned} \Delta E &=-\frac{mgR_E^2}{2r_2} -\left(-\frac{mgR_E^2}{2r_1}\right) \\ &=\frac{mgR_E^2}{2} \left( \frac{1}{r_1} - \frac{1}{r_2} \right). \end{aligned}$$

7. Replacing the radii,

$$\begin{aligned} \Delta E &=\frac{mgR_E^2}{2} \left( \frac{1}{1.5R_E} - \frac{1}{3R_E} \right) \\ &=\frac{mgR_E}{2} \left( \frac{2}{3} - \frac{1}{3} \right) \\ &=\frac{mgR_E}{6}. \end{aligned}$$

8. Using

$$m=100,\qquad g=10,\qquad R_E=6\times10^6\ \mathrm{m},$$

we obtain

$$\begin{aligned} \Delta E &=\frac{100\times10\times6\times10^6}{6} \\ &=1000\times10^6\ \mathrm{J}. \end{aligned}$$

9. Since

$$\Delta E=\alpha\times10^6\ \mathrm{J},$$

it follows that

$$\boxed{\alpha=1000}.$$

10. Therefore, Option B is correct.

1. According to Newton's law of universal gravitation, the force exerted by a mass $m$ on a test mass $m_0$ at the center is

$$F=\frac{Gmm_0}{r^2},$$

where $r$ is the distance from a corner of the square to its center.

2. Let

$$F_0=\frac{GMm_0}{r^2}.$$

Then the forces due to the four masses have magnitudes:

• $M \rightarrow F_0$
• $2M \rightarrow 2F_0$
• $3M \rightarrow 3F_0$
• $4M \rightarrow 4F_0$

3. Case 1: Original arrangement.

Along one diagonal, the forces due to $3M$ and $M$ act in opposite directions, giving

$$3F_0-F_0=2F_0.$$

Along the other diagonal, the forces due to $4M$ and $2M$ also oppose each other, giving

$$4F_0-2F_0=2F_0.$$

These two resultant forces are perpendicular, so

$$\begin{aligned} F_1 &=\sqrt{(2F_0)^2+(2F_0)^2} \\ &=2\sqrt{2}\,F_0. \end{aligned}$$

4. Case 2: After interchanging the positions of $3M$ and $4M$.

Along one diagonal, the net force becomes

$$3F_0-2F_0=F_0.$$

Along the other diagonal, the net force becomes

$$4F_0-F_0=3F_0.$$

Again, these two forces are perpendicular. Therefore,

$$\begin{aligned} F_2 &=\sqrt{(F_0)^2+(3F_0)^2} \\ &=\sqrt{10}\,F_0. \end{aligned}$$

5. Hence,

$$\begin{aligned} \frac{F_1}{F_2} &=\frac{2\sqrt{2}\,F_0}{\sqrt{10}\,F_0} \\ &=\frac{2}{\sqrt{5}}. \end{aligned}$$

6. Comparing with

$$\frac{F_1}{F_2} = \frac{\alpha}{\sqrt{5}},$$

we obtain

$$\boxed{\alpha=2}.$$

7. Therefore, Option A is correct.

1. Escape velocity is the minimum speed required for an object to escape a planet's gravitational field and reach infinity with zero final speed.

Using conservation of mechanical energy,

$$\frac{1}{2}mv_e^2-\frac{GMm}{R}=0.$$

Therefore,

$$v_e=\sqrt{\frac{2GM}{R}}.$$

2. The mass of a spherical planet is

$$M=\frac{4}{3}\pi R^3\rho,$$

where $\rho$ is the density of the planet.

3. Substituting this into the escape velocity formula,

$$\begin{aligned} v_e &=\sqrt{ \frac{2G}{R} \left( \frac{4}{3}\pi R^3\rho \right) } \\ &=\sqrt{ \frac{8}{3}\pi GR^2\rho }. \end{aligned}$$

Hence,

$$\begin{aligned} v_e &=R\sqrt{\frac{8}{3}\pi G\rho}. \end{aligned}$$

Therefore,

$$v_e\propto R\sqrt{\rho}.$$

4. For planet $A$,

$$v_{eA}=10\ \mathrm{km/s}=10000\ \mathrm{m/s}.$$

5. Planet $B$ has

$$R_B=0.1R_A, \qquad \rho_B=0.1\rho_A.$$

Using the proportionality,

$$\begin{aligned} \frac{v_{eB}}{v_{eA}} &=\frac{R_B\sqrt{\rho_B}} {R_A\sqrt{\rho_A}} \\ &=0.1\sqrt{0.1} \\ &=\frac{1}{10\sqrt{10}}. \end{aligned}$$

6. Therefore,

$$\begin{aligned} v_{eB} &=10000 \times \frac{1}{10\sqrt{10}} \\ &=\frac{1000}{\sqrt{10}} \\ &=100\sqrt{10}\ \mathrm{m/s}. \end{aligned}$$

7. Hence, the escape velocity from planet $B$ is

$$\boxed{100\sqrt{10}\ \mathrm{m/s}},$$

so Option D is correct.