Gravitation

W.D. $=U_2-U_1$

$ \begin{aligned} & =-\frac{G M m}{R+R}-\left(\frac{-G M m}{R}\right) \\ & =-\frac{G M m}{2 R}+\frac{G M m}{R} \\ & =\frac{G M m}{2 R}=\frac{m g R}{2} \end{aligned} $

To determine the length of one year on Mercury, we use Kepler's Third Law, which states that the square of the orbital period $ T $ is proportional to the cube of the mean radius $ R $ of the orbit:

$ T^2 \propto R^3 $

Given:

The radius of Mars' orbit around the Sun, $ R' = 4R $, where $ R $ is the radius of Mercury's orbit.

The Martian year $ T' = 687 $ Earth days.

First, apply Kepler's Third Law to find the ratio of the orbital periods:

$ \left(\frac{T'}{T}\right)^2 = \left(\frac{R'}{R}\right)^3 = \left(\frac{4R}{R}\right)^3 = 4^3 = 64 $

From this, we can solve for the ratio of the periods:

$ \frac{T'}{T} = 8 $

This means that Mars takes 8 times longer to orbit the Sun compared to Mercury. Therefore, the length of one year on Mercury is:

$ T = \frac{T'}{8} = \frac{687}{8} \approx 85.88 \text{ days} $

Hence, one year on Mercury is approximately 86 Earth days.

To find the gravitational force on a body at a height equal to one-third the Earth's radius from its surface, we start with the weight of the body on the Earth’s surface, which is 48 N.

The gravitational force at the surface, $ W $, is given by:

$ W = mg $

Where $ g = \frac{GM}{R^2} $.

At a height $ h $ above the Earth's surface, the gravitational force $ g_h $ is:

$ g_h = \frac{GM}{(R+h)^2} $

To find the weight $ W_h $ at this height, the ratio of gravitational forces at height $ h $ and at the surface is:

$ \frac{W_h}{W} = \frac{g_h}{g} = \frac{R^2}{(R+h)^2} $

Given $ h = \frac{R}{3} $, substitute $ h $ into the equation:

$ \frac{W_h}{W} = \frac{R^2}{\left(R+\frac{R}{3}\right)^2} = \frac{R^2}{\left(\frac{4R}{3}\right)^2} = \frac{9}{16} $

Thus, the new weight $ W_h $ is:

$ W_h = \frac{9}{16} \times W = \frac{9}{16} \times 48 \, \text{N} $

$ W_h = 27 \, \text{N} $

Therefore, at this height, the gravitational force experienced by the body is 27 N.

Escape velocity of object from planet is given by

$\begin{aligned} & \left(v_e\right)_p=\sqrt{\frac{2 G M_p}{R_p}} \\ & \therefore \quad\left(v_e\right)_p \propto \sqrt{\frac{M_p}{R_p}} \end{aligned}$

Now, $M_p=9 M_e$ and $R_\rho=16 R_e$ (given)

$\begin{aligned} & \frac{\left(v_e\right)_p}{\left(v_e\right)_e}=\sqrt{\frac{M_p}{R_p} \times \frac{R_e}{M_e}}=\sqrt{\frac{9 M_e}{16 R_e} \times \frac{R_e}{M_e}}=\sqrt{\frac{9}{16}}=\frac{3}{4} \\ & \Rightarrow \quad\left(v_e\right)_p=\frac{3}{4} v \end{aligned}$

$M g^{\prime}=M g \frac{R^2}{(R+h)^2}$

At $A$ $\quad M g^{\prime}=M g \frac{R^2}{(R+2 R)^2}=\frac{M g}{9}$

At $B \quad M g^{\prime}=M g \frac{R}{\left(R+\frac{3 R}{2}\right)^2}=\frac{M g \cdot 4}{25}$

Change in weight $=M g \frac{4}{25}-\frac{M g}{9}=49 \mathrm{~N}$

The acceleration due to gravity (g) on a planet is given by the formula:

$ g = G \frac{M}{R^2} $

where:

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

In this question, we know that:

• The mass of the planet $ M_p $ is $\frac{1}{10}$ of the mass of the earth $ M_e $, so $ M_p = \frac{M_e}{10} $.
• The diameter of the planet is half that of earth. Since the diameter is twice the radius, a diameter half that of earth implies the radius $ R_p $ is also half the radius of the earth $ R_e $. Therefore, $ R_p = \frac{R_e}{2} $.

Using the formula for acceleration due to gravity and substituting the above information:

$ g_p = G \frac{M_p}{R_p^2} = G \frac{\frac{M_e}{10}}{\left(\frac{R_e}{2}\right)^2} $

Simplify the expression:

$ g_p = G \frac{M_e}{10} \cdot \frac{4}{R_e^2} = \frac{4}{10} G \frac{M_e}{R_e^2} $

Knowing that $ G \frac{M_e}{R_e^2} $ is the acceleration due to gravity on Earth $ g_e \approx 9.8 \, \mathrm{m/s}^2 $, we substitute this value into the equation:

$ g_p = \frac{4}{10} \cdot 9.8 \, \mathrm{m/s}^2 = 3.92 \, \mathrm{m/s}^2 $

Thus, the acceleration due to gravity on the planet is $3.92 \, \mathrm{m/s}^2$, which corresponds to:

Option D: $3.92 \mathrm{~m/s}^2$

Apply energy conservation,

$\begin{aligned} & U_i+K_i=U_f+K_f \\ & \Rightarrow \quad-\frac{G M m}{R}+K_i=-\frac{G M m}{3 R}+\frac{1}{2} m v^2 \\ & \Rightarrow \quad-\frac{G M m}{R}+K_i=-\frac{G M m}{3 R}+\frac{1}{2} \times m \times \frac{G M}{3 R} \\ & \Rightarrow \quad K_i=-\frac{1}{6} \frac{G M m}{R}+\frac{G M m}{R} \\ & K_i=\frac{5}{6} \frac{G M m}{R} \end{aligned}$

$V_{\infty}=\sqrt{V^2-V_e^2}$

Given than

$\mathrm{V}=2 \mathrm{~V}_e$

So,

$\begin{aligned} & \mathrm{V}_{\infty}=\sqrt{\left(2 \mathrm{~V}_e\right)^2-\mathrm{V}_e^2} \\ & \mathrm{~V}_{\infty}=\sqrt{3} \mathrm{~V}_e=11.2 \sqrt{3} \mathrm{~km} / \mathrm{s}\end{aligned}$

$ \begin{aligned} & g=\frac{4}{3} \pi \mathrm{GR} \rho \\ & \rho=\frac{3 g}{4 \pi \mathrm{GR}} \end{aligned} $

Position of Neutral point (Zero Gravitational Field)

$r_{1}=\frac{\sqrt{m_{1}} R}{\sqrt{m_{1}}+\sqrt{m_{2}}}=\frac{\sqrt{m} R}{\sqrt{m}+\sqrt{9 m}}=\frac{R}{4}$

$\mathrm{r}_{2}=\mathrm{R}-\mathrm{R} / 4=3 \mathrm{R} / 4$

Now Gravitational potential at point $\mathrm{P}$

$ \begin{aligned} & V_{P}=-\frac{G M}{R / 4}-\frac{9(\mathrm{GM})}{3 \mathrm{R} / 4} \\ & =\frac{-16 \mathrm{GM}}{\mathrm{R}} \end{aligned} $

For a satellite orbiting just above the surface of the Earth, we can use the formula for the period T of the satellite in terms of the Earth's density d and the gravitational constant G:

$T = 2\pi\sqrt{\frac{a^3}{GM}}$

where a is the semi-major axis of the orbit (which is approximately equal to the Earth's radius R for a satellite orbiting just above the surface) and M is the mass of the Earth.

We can express the mass of the Earth M in terms of its density d and volume:

$M = dV = d\times\frac{4}{3}\pi R^3$

Now, substitute this expression for M into the equation for T:

$T = 2\pi\sqrt{\frac{a^3}{Gd\times\frac{4}{3}\pi R^3}}$

Since the satellite is orbiting just above the Earth's surface, we can approximate a ≈ R:

$T = 2\pi\sqrt{\frac{R^3}{Gd\times\frac{4}{3}\pi R^3}}$

Simplify the equation:

$T = 2\pi\sqrt{\frac{1}{Gd\times\frac{4}{3}\pi}}$

Now, square both sides of the equation:

$T^2 = 4\pi^2\frac{1}{Gd\times\frac{4}{3}\pi}$

Simplify further:

$T^2 = \frac{3\pi}{Gd}$

Thus, the quantity $\frac{3\pi}{Gd}$ represents $T^2$.

Since the gravitational field is conservative in nature hence the work done would depend only on the initial and final positions and not on the path followed by the mass.

Hence, W₁ = W₂ = W₃

Gravitational field intensity $ = {{ - \partial V} \over {\partial x}}\widehat i + {{ - \partial V} \over {\partial y}}\widehat j + {{ - \partial V} \over {\partial x}}\widehat k$

where V is gravitational potential

$\Rightarrow$ Gravitational field intensity $ = {{ - \partial } \over {\partial x}}\left[ {{{ - K} \over x}} \right] = {{ - K} \over {{x^2}}}\widehat i$

at (2, 0, 3)

$\overrightarrow {{E_G}} = {{ - K} \over {{2^2}}} = {{ - K} \over 4}\widehat i$

$F = m{E_G}$

$3 = {{60} \over {1000}}{E_G}$

${E_G} = 50$ N/kg

(a) $[G] = {{F{r^2}} \over {{m_1}{m_2}}}$

$[G] = {{F{r^2}} \over {{m_1}{m_2}}} = {{[ML{T^{ - 2}}]{L^2}} \over {[MM]}} = [{M^{ - 1}}{L^3}{T^{ - 2}}]$

(b) Gravitational potential energy $ = [M{L^2}{T^{ - 2}}]$

(c) Gravitational potential $ = {{PE} \over m} = [{L^2}{T^{ - 2}}]$

(d) Gravitational field intensity $ = {F \over m} = [L{T^{ - 2}}]$

Given, radius of earth, $R_e=6400 \mathrm{~km}$

Distance from centre of earth, $r=2000 \mathrm{~km}$

Distance from the surface of earth,

$\begin{aligned} d & =R_e-r \\ & =6400-2000 \\ & =4400 \mathrm{~km}=4.4 \times 10^6 \mathrm{~m} \end{aligned}$

If $g^{\prime}$ be the gravitation field below the surface of earth at a depth $d$, then

$\begin{aligned} g^{\prime} & =g\left(1-\frac{d}{R_{\mathrm{e}}}\right)=9.8\left(1-\frac{4.4 \times 10^6}{6.4 \times 10^6}\right) \quad\left[\because g=9.8 \mathrm{~m} / \mathrm{s}^2\right] \\ & =9.8 \times 0.3125=3.06 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$

From Kepler's law,

$\begin{array}{ll} & \frac{T_A^2}{T_B^2}=\frac{r_A^3}{r_B^3} \Rightarrow \frac{1^2}{8^2}=\frac{\left(10^4\right)^3}{r_B^3} \\ \Rightarrow & r_B^3=64 \times\left(10^4\right)^3 \\ \therefore & r_B=4 \times 10^4 \mathrm{~km} \end{array}$

Speed of satellite $A$,

$v_A=\frac{2 \pi r_A}{T_A}=\frac{2 \pi \times 10^4}{1}$

$=2 \pi \times 10^4 \mathrm{~km} / \mathrm{h}$

Speed of satellite B,

$\begin{aligned} & v_B=\frac{2 \pi r_B}{T_B}\\ & =\frac{2 \pi \times 4 \times 10^4}{8} \\ & =\pi \times 10^4 \mathrm{~km} / \mathrm{h} \end{aligned}$

The speed of $B$ relative to $A$ when they are closed.

$\begin{aligned} v_{B A} & =v_A-v_B \\ & =2 \pi \times 10^4-\pi \times 10^4 \\ & =\pi \times 10^4 \mathrm{~km} / \mathrm{h} \end{aligned}$

As, force $=\frac{G M m}{r^{3 / 2}}=m \omega^2 r$

$\begin{aligned} & \Rightarrow \quad \frac{G M m}{r^{3 / 2}}=\frac{4 \pi^2 m r}{T^2} \quad\left[\because T=\frac{2 \pi}{\omega}\right] \\ & \Rightarrow \quad T^2=\left(\frac{\Delta \pi^2}{G M}\right) \cdot r^{5 / 2} \\ & \Rightarrow \quad T^2 \propto r^{5 / 2} \\ \end{aligned}$

Given, $h=\frac{R_e}{2}$

Acceleration due to gravity at altitude $h$ is given by

$\begin{aligned} g^{\prime} & =\frac{g}{\left(1+\frac{h}{R_e}\right)^2}=\frac{g}{\left(1+\frac{R_e / 2}{R_e}\right)^2} \\ & =\frac{g}{\left(1+\frac{1}{2}\right)^2}=\frac{g}{\left(\frac{3}{2}\right)^2}=\frac{4 g}{9} \quad \text{.... (i)} \end{aligned}$

Weight of the body at the earth's surface,

$w=m g=63 \mathrm{~N}$ .... (ii)

Weight of the body at altitude, $h=\frac{R_e}{2}$

$w^{\prime}=m g^{\prime}=\frac{4}{9} m g$ ..... (iii)

Using Eq. (ii), we get

$w^{\prime}=\frac{4}{9} \times 63=28 \mathrm{~N}$

The orbital velocity of space ship in circular

$v_0=\sqrt{\frac{G M}{r}}$

If space ship is very close to earth surface, r = radius of earth = R

$\begin{aligned} & \therefore \quad v_0=\sqrt{\frac{G M}{R}}=\sqrt{\frac{R^2 g}{R}}=\sqrt{R g} \\ &\left(\text { since } g=\frac{G M}{R^2}\right) \end{aligned}$

$\begin{aligned} \text { Here, } \quad R & =6400 \mathrm{~km}=6.4 \times 10^6 \mathrm{~m} \\ g & =9 \cdot 8 \mathrm{~m} / \mathrm{s}^2 \\ \therefore \quad v_0 & =\sqrt{6 \cdot 4 \times 10^6 \times 9 \cdot 8} \\ & =7.92 \times 10^3 \mathrm{~m} / \mathrm{s} \\ & =7.92 \mathrm{~km} / \mathrm{s} \end{aligned}$

The escape velocity of space ship

$\begin{aligned} v_e & =\sqrt{(2 R g)}=\sqrt{2} v_0=\sqrt{2} \times 7 \cdot 92 \\ & =11 \cdot 20 \mathrm{~km} / \mathrm{s} \end{aligned}$

$\therefore$ Additional velocity required

$=11 \cdot 20-7 \cdot 92=3 \cdot 28 \mathrm{~km} / \mathrm{s}$

Therefore, velocity $3.28 \mathrm{~km} / \mathrm{s}$ must be added to orbital velocity to spaceship to overcome the gravitational pull.

From conservation of energy,

$\frac{1}{2} m v_1^2-\frac{G M_e m}{R}=\frac{1}{2} m v_2^2-\frac{G M_e m}{R+H}$

Here, $v_2=0$ (At maximum height)

$\therefore \quad \frac{1}{2} m v_1-\frac{G M_e m}{R}=-\frac{G M_e m}{R+H}$

Solving, we get

$H=\frac{R}{\frac{2 g R}{v_1^2}-1}=\frac{R}{\frac{v_e^2}{v_1^2}-1}$

Where $v_e$ is escape velocity

Here $v=\frac{1}{3} v_e \Rightarrow \frac{v_e}{v}=3$

$\therefore \quad H=\frac{R}{9-1}=\frac{R}{8}$

If $r_1$ and $r_2$ are radii of orbits of satellites $S_1$ and $S_2$ and $T_1, T_2$ are their respective time periods of revolution, then from Kepler's third law,

$\frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3} \Rightarrow r_2=\left(\frac{T_2}{T_1}\right)^{2 / 3} \times r_1$

$\begin{aligned} \text { Here, } \quad T_1 & =1 \mathrm{~h} \\ T_2 & =8 \mathrm{~h} \\ r_1 & =10^4 \mathrm{~km} \\ \Rightarrow \quad\left(\frac{8 h}{1 h}\right)^{2 / 3} \times 10^4 & =4 \times 10^4 \mathrm{~km} \end{aligned}$

The speed of satellite $S_2$ relative to $S_1$

$\begin{aligned} \left|v_2-v_1\right| & =\frac{2 \pi r_2}{T_2}-\frac{2 \pi r_1}{T_1} \\ & =2 \pi\left[\frac{r_2}{T_2}-\frac{r_1}{T_1}\right] \\ & =2 \pi\left[\frac{4 \times 10^4}{8}-\frac{10^4}{1}\right] \mathrm{km} / \mathrm{h} \\ & =2 \pi \times \frac{1}{2} \times 10^4 \\ & =\pi \times 10^4 \mathrm{~km} / \mathrm{h} \end{aligned}$