Gravitation

• Escape velocity v = √(2GM/R).
• Earth: v = √(2GMₑ/Rₑ). Planet: M' = 9Mₑ, R' = 16Rₑ. v' = √(2G(9Mₑ)/(16Rₑ)) = √(9/16) * √(2GMₑ/Rₑ) = (3/4)v.

(NEW NCERT 11^th Page No. 135, 136)

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: $$R_{\text{Planet}} = 2 R_{\text{Earth}}$$​

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

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

$$\boxed{22.4 \text{ km/s}}$$​

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

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

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

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

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

$T_2=1day$

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:

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

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

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

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

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

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

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

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) $-\frac{2\gamma }{m_2}$

key concept= Columb's low + Law of gravitation

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

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

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

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 $A$.

• 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) $-\frac{2\gamma }{m_2}$

Escape velocity:

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

Given:

$$M_m=\frac{M}{144}, \quad R_m=\frac{R}{16}$$

$$\frac{v_m}{v}=\sqrt{\frac{M_m/R_m}{M/R}} =\sqrt{\frac{(M/144)}{(R/16)}\cdot \frac{R}{M}}$$

$$=\sqrt{\frac{16}{144}}=\frac{1}{3}$$

$$v_m=\frac{v}{3}$$