Gravitation

The formula for escape velocity is:

From the given problem, the ratios for the moon compared to the planet are:

Mass: $M_m=\frac{M_p}{144}$

Radius: $R_m=\frac{R_p}{16}$ (the ratio of radii is the same as the ratio of diameters)

Now, we can find the escape velocity on the moon ($v_m$) by setting up a ratio:

Substitute the given values:

Correct Option: D

The formula for escape velocity is:

From the given problem, the ratios for the moon compared to the planet are:

Mass: $M_m=\frac{M_p}{144}$

Radius: $R_m=\frac{R_p}{16}$ (the ratio of radii is the same as the ratio of diameters)

Now, we can find the escape velocity on the moon ($v_m$) by setting up a ratio:

Substitute the given values:

Correct Option: D

For a uniform-density Earth, acceleration due to gravity at a depth d is

g_d=g\left(1-\frac{d}{R}\right)

Given:

$$d=\frac{R}{4}$$

Therefore,

$$g_d=g(1-\frac{1}{4})$$ $$g_d=\frac{3}{4}g$$

Weight is proportional to g.

$$W_d=300\times \frac{3}{4}$$ $$W_d=225\text{N}$$

Using Kepler’s third law:

$T\propto \frac{R^{3/2}}{\sqrt{M}}$

For $P_1:T_1\propto \frac{R^{3/2}}{\sqrt{2M}}$
For $P_2:T_2\propto \frac{(2R{)}^{3/2}}{\sqrt{4M}}=\frac{\sqrt{2}{R}^{3/2}}{\sqrt{M}}$
Ratio: $\frac{T_2}{T_1}=\frac{\sqrt{2}/\sqrt{M}}{1/\sqrt{2M}}=2$

$\frac{dA}{dt}=\frac{L}{2m}$
Due to central force torque is zero & angular momentum is constant.

key concept= Columb's low + Law of gravitation

$F_e=\frac{k{r}_1{r}_2}{r^2}=\frac{9\times {10}^9\times {\left(1.6\times {10}^{-19}\right)}^2}{r^2}$

$F_g=\frac{G{m}_1{m}_2}{r^2}=\frac{6.67\times {10}^{-11}\times 9.1\times {10}^{-31}\times 1.67\times {10}^{-27}}{r^2}$

$\therefore \frac{{\mathrm{F}}_{\mathrm{e}}}{{\mathrm{F}}_{\mathrm{g}}}=2.2\times {10}^{39}$

• System: Binary star system with masses $m_1$and$m_2$ ($m_1\gg m_2$).
• Process: Mass transfer from lighter star ($m_2$) to heavier star ($m_1$) at a rate $\gamma$.
  • ${\dot{m}}_1=\gamma$
  • ${\dot{m}}_2=-\gamma$
  • Total mass $M=m_1+m_2$is constant.
• Principle: Conservation of orbital angular momentum $L$.

Step 1: Write the expression for Angular Momentum. The orbital angular momentum $L$ for a binary system is given by: $L=\mu \sqrt{GMr}$ where $\mu =\frac{m_1{m}_2}{m_1+m_2}$ is the reduced mass, $M$ is the total mass, and $r$ is the separation.

Step 2: Differentiate with respect to time. Taking the natural logarithm of both sides: $\text{ln}L=\text{ln}\mu +\frac{1}{2}\text{ln}\left(GM\right)+\frac{1}{2}\text{ln}r$ Since $L$,$G$, and $M$ are constant, their derivatives are zero. Differentiating with respect to time $t$ : $0=\frac{1}{\mu }\frac{d\mu }{dt}+\frac{1}{2r}\frac{dr}{dt}$Rearranging to solve for the relative rate of change of r : $\frac{1}{r}\frac{dr}{dt}=-2\frac{1}{\mu }\frac{d\mu }{dt}$

Step 3: Calculate $\frac{1}{\mu }\frac{d\mu }{dt}$. Using the definition of reduced mass $\mu =\frac{m_1{m}_2}{M}$: $\text{ln}\mu =\text{ln}{m}_1+\text{ln}{m}_2-\text{ln}M$ $\frac{1}{\mu }\frac{d\mu }{dt}=\frac{{\dot{m}}_1}{m_1}+\frac{{\dot{m}}_2}{m_2}$Substitute ${\dot{m}}_1=\gamma$ and ${\dot{m}}_2=-\gamma$: $\frac{1}{\mu }\frac{d\mu }{dt}=\frac{\gamma }{m_1}-\frac{\gamma }{m_2}=\gamma \left(\frac{1}{m_1}-\frac{1}{m_2}\right)$

Step 4: Apply the approximation $m_1\gg m_2$. Since $m_1$ is much larger than $m_2$, the term $\frac{1}{m_1}$is negligible compared to $\frac{1}{m_2}.\frac{1}{\mu }\frac{d\mu }{dt}\approx \gamma \left(0-\frac{1}{m_2}\right)=-\frac{\gamma }{m_2}$

Step 5: Final Result. Substitute this back into the equation for $\frac{1}{r}\frac{dr}{dt}:\frac{1}{r}\frac{dr}{dt}=-2\left(-\frac{\gamma }{m_2}\right)=\frac{2\gamma }{m_2}$

Note: While the derived physical result is positive (indicating orbit expansion), the options provided in the question all have a negative sign. This suggests a sign convention difference or a typo in the question statement regarding the direction of mass flow (usually Heavy $\to$ Light causes shrinking/negative rate). However, the magnitude and dependency on $m_2$ clearly point to Option (B).

Answer: (B) $-\frac{2\gamma }{m_2}$

$\frac{\mathrm{GMm}}{{\mathrm{r}}_0^2}={\mathrm{m\omega }}^2{\mathrm{r}}_0$

$\mathrm{\omega }=\sqrt{\frac{\mathrm{GM}}{{\mathrm{r}}_0^3}}$

${\mathrm{T}}_0^2=4{\mathrm{\pi }}^2\left(\frac{{\mathrm{r}}_0^3}{\mathrm{GM}}\right)$ .....(1)

$\mathrm{F}=-\frac{{\mathrm{dv}}_{\mathrm{c}}\left(\mathrm{r}\right)}{\mathrm{dr}}$

$\mathrm{F}=\frac{3\mathrm{m\alpha }}{{\mathrm{r}}^4}$

Net force

$\frac{\mathrm{GMm}}{{\mathrm{r}}_0^2}-\frac{3\mathrm{m\alpha }}{{\mathrm{r}}_0^4}={\mathrm{m\omega }}^2{\mathrm{r}}_0$

${\mathrm{\omega }}^2=\frac{\mathrm{GM}}{{\mathrm{r}}_0^3}-\frac{3\mathrm{\alpha }}{{\mathrm{r}}_0^5}$

We know $\mathrm{\omega }=\frac{2\mathrm{\pi }}{{\mathrm{T}}_1}$

$\frac{4{\mathrm{\pi }}^2}{{\mathrm{T}}_1^2}=\frac{\mathrm{GM}}{{\mathrm{r}}_0^3}-\frac{3\mathrm{\alpha }}{{\mathrm{r}}_0^5}$

${\mathrm{T}}_1^2=\frac{4{\mathrm{\pi }}^2}{\frac{\mathrm{GM}}{{\mathrm{r}}_0^3}-\frac{3\mathrm{\alpha }}{{\mathrm{r}}_0^5}}$ ......(2)

From equation (1) and (2)

$\frac{{\mathrm{T}}_1^2-{\mathrm{T}}_0^2}{{\mathrm{T}}_1^2}=\frac{3\mathrm{\alpha }}{{\mathrm{GMr}}_0^2}$

Initial potential energy at Earth’s surface:
$U_1=-\frac{GMm}{R_e}$

Final height is $2R_e$ above the surface, so final distance from Earth’s center is $3R_e$:
$U_2=-\frac{GMm}{3R_e}$
Increase in potential energy:
$\Delta U=U_2-U_1$
$=-\frac{GMm}{3R_e}+\frac{GMm}{R_e}=\frac{2GMm}{3R_e}$
Using
$g=\frac{GM}{R_e^2}\Rightarrow GM=g{R}_e^2$
$\Delta U=\frac{2}{3}mg{R}_e$

$g^{\text{'}}=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$

$\frac{g}{9}=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$

${\left(1+\frac{h}{R}\right)}^2=9$

$1+\frac{h}{R}=3$

$\frac{h}{R}=2$

$\therefore h=2R$

(b) Acceleration due to gravity at a depth

(d) below the surface is $g_d=g\left(1-\frac{d}{R}\right)$

At a height (h) above the surface is
$g_h=g\left(1-\frac{2h}{R}\right)$

Here d = h = 16 km and R = 6400 km.
Thus,
$g_d=g\left(1-\frac{16}{6400}\right)=g\left(1-\frac{1}{400}\right)$

$g_d=g\left(1-\frac{16}{6400}\right)=g\left(1-\frac{1}{400}\right)$

$\Delta g=g_d-g_h=g\left(\frac{1}{200}-\frac{1}{400}\right)=\frac{g}{400}$

$\frac{\Delta g}{g}\times 100=\frac{1}{400}\times 100=0.25\%$

By Kepler's Third Law, $T \propto R^{3/2}$.

• Time period ratio: $$\frac{T_B}{T_A} = \left( \frac{R_B}{R_A} \right)^{3/2} = (1.03)^{3/2}.$$
• Using binomial approximation, $(1.03)^{3/2} \approx 1 + \frac{3}{2}(0.03) = 1.045$
• Percentage increase: $$\Delta T \% = (1.045 - 1) \times 100 = 4.5\%.$$

Thus, the time period of $B$ is approximately 4.5% greater than that of $A$.

The angular momentum ($L$) of a planet revolving around a massive star in a circular orbit is given by:

$$L = m r v$$

From Kepler’s laws, the orbital velocity $v$ is given by:

$$v = \sqrt{\frac{GM}{r}}$$

Substituting this into the angular momentum equation:

$$L = m r \sqrt{\frac{GM}{r}}$$

​ $$L = m \sqrt{G M r}$$​

For planets A and B, we have:

• $r_A = 2r_B$​
• $m_A = 4\sqrt{3} m_B$​

Since $L$ depends on $m$ and $\sqrt{r}$​:

$$L_A = m_A \sqrt{G M r_A} = (4\sqrt{3} m_B) \sqrt{G M (2r_B)}$$

​ $$= 4\sqrt{3} m_B \times \sqrt{2} \times \sqrt{G M r_B}$$

​ $$= 4\sqrt{6} m_B \sqrt{G M r_B}$$

Similarly, for planet B:

$$L_B = m_B \sqrt{G M r_B}$$

$$\frac{L_B}{L_A} = \frac{m_B \sqrt{G M r_B}}{4\sqrt{6} m_B \sqrt{G M r_B}}$$​​ $$= \frac{1}{4\sqrt{6}}$$​

$$\boxed{B} \quad \frac{1}{4\sqrt{6}}$$​

Fime-Period Relatien: The time period T of a satellite in orbit is related to the radius R of its orbit by the equation:

$T^2\propto R^3$

Therefore,

$\frac{T_2^2}{T_1^2}=\frac{(1.03R{)}^3}{R^3}$

Simplifying:

$\frac{T_2}{T_1}=(1.03{)}^{3/2}\approx 1.045$

Therefore, the increase in the time period is approximately:

$\frac{\Delta T}{T}=\frac{1.045-1}{1}\times 100=4.5\%$

$$\begin{gathered} T^2\propto R^3 \\ {\left(\frac{T_1}{T_2}\right)}^2={\left(\frac{R_1}{R_2}\right)}^3 \\ T_1=27\text{ days } R_1=R R_2=R/9 \\ {\left(\frac{27}{T_2}\right)}^2={\left(\frac{R}{R/9}\right)}^3 \\ T_2^2=\frac{27\times 27}{9\times 9\times 9} \\ T_2^2=\frac{3\times 3}{9} \end{gathered}$$

$T_2=1day$

$$\begin{gathered} T^2\propto R^3 \\ {\left(\frac{T_1}{T_2}\right)}^2={\left(\frac{R_1}{R_2}\right)}^3 \\ T_1=27\text{ days } R_1=R R_2=R/9 \\ {\left(\frac{27}{T_2}\right)}^2={\left(\frac{R}{R/9}\right)}^3 \\ T_2^2=\frac{27\times 27}{9\times 9\times 9} \\ T_2^2=\frac{3\times 3}{9} \end{gathered}$$

$T_2=1day$

Escape velocity ($v_e$​) is given by:

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

where:

• $G$ is the gravitational constant.
• $M$ is the mass of the planet.
• $R$ is the radius of the planet.

For Earth:

$$v_{e, \text{Earth}} = 11.2 \text{ km/s}$$

The escape velocity of a planet compared to Earth is:

$$\frac{v_{e, \text{Planet}}}{v_{e, \text{Earth}}} = \sqrt{\frac{M_{\text{Planet}} / M_{\text{Earth}}}{R_{\text{Planet}} / R_{\text{Earth}}}}$$

• The mass of the planet is 8 times the mass of Earth: $$M_{\text{Planet}} = 8 M_{\text{Earth}}$$​
• The radius of the planet is 2 times the radius of Earth: $$R_{\text{Planet}} = 2 R_{\text{Earth}}$$​

$$\frac{v_{e, \text{Planet}}}{v_{e, \text{Earth}}} = \sqrt{\frac{8M_{\text{Earth}}}{2R_{\text{Earth}}}}$$​ $$= \sqrt{\frac{8}{2}} = \sqrt{4} = 2$$

​ $$v_{e, \text{Planet}} = 2 \times 11.2 = 22.4 \text{ km/s}$$

$$\boxed{22.4 \text{ km/s}}$$​

Conservation of Angular Momentum

For conservative mass transfer, the total orbital angular momentum (\(L\)) remains constant.

(where total mass $M=m_1+m_2$ is constant).

Logarithmic Differentiation

Taking the natural logarithm of both sides to easily separate the variables:

Differentiating with respect to time \(t\):

Apply Given Conditions

The heavier star gains mass ($\frac{d{m}_1}{dt}=\gamma$) and the lighter star loses mass ($\frac{d{m}_2}{dt}=-\gamma$). Substitute these rates:

Since $m_1\gg m_2$, the term $\frac{\gamma }{m_1}$ is negligibly small compared to $\frac{\gamma }{m_2}$:

Final Rearrangement

Correct Option: B

$\frac{\mathrm{GMm}}{{\mathrm{r}}_0^2}={\mathrm{m\omega }}^2{\mathrm{r}}_0$

$\mathrm{\omega }=\sqrt{\frac{\mathrm{GM}}{{\mathrm{r}}_0^3}}$

${\mathrm{T}}_0^2=4{\mathrm{\pi }}^2\left(\frac{{\mathrm{r}}_0^3}{\mathrm{GM}}\right)$ .....(1)

$\mathrm{F}=-\frac{{\mathrm{dv}}_{\mathrm{c}}\left(\mathrm{r}\right)}{\mathrm{dr}}$

$\mathrm{F}=\frac{3\mathrm{m\alpha }}{{\mathrm{r}}^4}$

Net force

$\frac{\mathrm{GMm}}{{\mathrm{r}}_0^2}-\frac{3\mathrm{m\alpha }}{{\mathrm{r}}_0^4}={\mathrm{m\omega }}^2{\mathrm{r}}_0$

${\mathrm{\omega }}^2=\frac{\mathrm{GM}}{{\mathrm{r}}_0^3}-\frac{3\mathrm{\alpha }}{{\mathrm{r}}_0^5}$

We know $\mathrm{\omega }=\frac{2\mathrm{\pi }}{{\mathrm{T}}_1}$

$\frac{4{\mathrm{\pi }}^2}{{\mathrm{T}}_1^2}=\frac{\mathrm{GM}}{{\mathrm{r}}_0^3}-\frac{3\mathrm{\alpha }}{{\mathrm{r}}_0^5}$

${\mathrm{T}}_1^2=\frac{4{\mathrm{\pi }}^2}{\frac{\mathrm{GM}}{{\mathrm{r}}_0^3}-\frac{3\mathrm{\alpha }}{{\mathrm{r}}_0^5}}$ ......(2)

From equation (1) and (2)

$\frac{{\mathrm{T}}_1^2-{\mathrm{T}}_0^2}{{\mathrm{T}}_1^2}=\frac{3\mathrm{\alpha }}{{\mathrm{GMr}}_0^2}$

2

key concept= Coulomb's Law + Law of gravitation

$F_e=\frac{k{r}_1{r}_2}{r^2}=\frac{9\times {10}^9\times {\left(1.6\times {10}^{-19}\right)}^2}{r^2}$

$F_g=\frac{G{m}_1{m}_2}{r^2}=\frac{6.67\times {10}^{-11}\times 9.1\times {10}^{-31}\times 1.67\times {10}^{-27}}{r^2}$

$\therefore \frac{{\mathrm{F}}_{\mathrm{e}}}{{\mathrm{F}}_{\mathrm{g}}}=2.2\times {10}^{39}$

Given kinetic energy expression written in the assertion.
$\mathrm{KE}=\frac{1\times m}{2}\times \left(\frac{2Gm}{R}\right)$
Comparing with gravitational potential energy term used in the assertion.
$\mathrm{KE}=mgR$
This equality is incorrect because gravitational potential energy is not defined as m g R in this context.
At infinite distance, gravitational potential energy is taken as zero.
$U=0$
Hence the assertion is incorrect.
The reason stating potential energy is zero at infinity is correct.
Final conclusion: assertion is wrong but reason is correct.

\begin{aligned}
& g=\frac{G M}{R^2} \Rightarrow g \propto \frac{1}{R^2} \\
& \frac{g_2}{g_1}=\frac{R_1^2}{R_2^2} \\
& g_2=4 g_1\left(R_2=\frac{R_1}{2}\right)
\end{aligned}

\begin{aligned}
& g=\frac{G M}{R^2} \Rightarrow g \propto \frac{1}{R^2} \\
& \frac{g_2}{g_1}=\frac{R_1^2}{R_2^2} \\
& g_2=4 g_1\left(R_2=\frac{R_1}{2}\right)
\end{aligned}

Acceleration due to gravity is:

If the diameter becomes half, then radius also becomes half:

Mass remains constant.

New acceleration due to gravity:

Substitute the new radius:

Angular speed is inversely proportional to time period
$\omega \propto \frac{1}{T}$
Time period of revolution of moon
$T_{\mathrm{moon}}=365+\frac{4}{24}$
Comparison of angular speeds
${\omega }_{\mathrm{moon}}>{\omega }_{\mathrm{earth}}$

Angular speed is inversely proportional to time period
$\omega \propto \frac{1}{T}$
Time period of revolution of moon
$T_{\mathrm{moon}}=365+\frac{4}{24}$
Comparison of angular speeds
${\omega }_{\mathrm{moon}}>{\omega }_{\mathrm{earth}}$

By Kepler's Third Law, $T \propto R^{3/2}$.

• Time period ratio: $$\frac{T_B}{T_A} = \left( \frac{R_B}{R_A} \right)^{3/2} = (1.03)^{3/2}.$$
• Using binomial approximation, $(1.03)^{3/2} \approx 1 + \frac{3}{2}(0.03) = 1.045$
• Percentage increase: $$\Delta T \% = (1.045 - 1) \times 100 = 4.5\%.$$

Thus, the time period of $B$ is approximately 4.5% greater than that of $A$.

Escape velocity ($v_e$​) is given by:

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

where:

• $G$ is the gravitational constant.
• $M$ is the mass of the planet.
• $R$ is the radius of the planet.

For Earth:

$$v_{e, \text{Earth}} = 11.2 \text{ km/s}$$

The escape velocity of a planet compared to Earth is:

$$\frac{v_{e, \text{Planet}}}{v_{e, \text{Earth}}} = \sqrt{\frac{M_{\text{Planet}} / M_{\text{Earth}}}{R_{\text{Planet}} / R_{\text{Earth}}}}$$

• The mass of the planet is 8 times the mass of Earth: $$M_{\text{Planet}} = 8 M_{\text{Earth}}$$​
• The radius of the planet is 2 times the radius of Earth: $$R_{\text{Planet}} = 2 R_{\text{Earth}}$$​

$$\frac{v_{e, \text{Planet}}}{v_{e, \text{Earth}}} = \sqrt{\frac{8M_{\text{Earth}}}{2R_{\text{Earth}}}}$$​ $$= \sqrt{\frac{8}{2}} = \sqrt{4} = 2$$

​ $$v_{e, \text{Planet}} = 2 \times 11.2 = 22.4 \text{ km/s}$$

$$\boxed{22.4 \text{ km/s}}$$​

The angular momentum ($L$) of a planet revolving around a massive star in a circular orbit is given by:

$$L = m r v$$

From Kepler’s laws, the orbital velocity $v$ is given by:

$$v = \sqrt{\frac{GM}{r}}$$

Substituting this into the angular momentum equation:

$$L = m r \sqrt{\frac{GM}{r}}$$

​ $$L = m \sqrt{G M r}$$​

For planets A and B, we have:

• $r_A = 2r_B$​
• $m_A = 4\sqrt{3} m_B$​

Since $L$ depends on $m$ and $\sqrt{r}$​:

$$L_A = m_A \sqrt{G M r_A} = (4\sqrt{3} m_B) \sqrt{G M (2r_B)}$$

​ $$= 4\sqrt{3} m_B \times \sqrt{2} \times \sqrt{G M r_B}$$

​ $$= 4\sqrt{6} m_B \sqrt{G M r_B}$$

Similarly, for planet B:

$$L_B = m_B \sqrt{G M r_B}$$

$$\frac{L_B}{L_A} = \frac{m_B \sqrt{G M r_B}}{4\sqrt{6} m_B \sqrt{G M r_B}}$$​​ $$= \frac{1}{4\sqrt{6}}$$​

$$\boxed{B} \quad \frac{1}{4\sqrt{6}}$$​