Numerical Problems-2 - Chapter 6 - Gravitation - 11th Science Guide Samacheer Kalvi Solutions

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Numerical Problems
Question 1.
A geo-stationary satellite is orbiting the Earth of a height of $6 \mathrm{R}$ above the surface of Earth $\mathrm{R}$ being the radius of the Earth calculate the time period of another satellite at a height of $2.5 \mathrm{R}$ from the surface of Earth.
Distance of satellite from the center are $7 \mathrm{R}$ and $3.5 \mathrm{R}$ respectively.
Answer:
$
\begin{aligned}
\frac{\mathrm{T}_2}{\mathrm{~T}_1} & =\left(\frac{a_2}{a_1}\right)^{\frac{3}{2}} \\
\mathrm{~T}_2 & =\mathrm{T}_1\left(\frac{a_2}{a_1}\right)^{\frac{3}{2}}=24\left(\frac{3.5 \mathrm{R}}{7 \mathrm{R}}\right)^{\frac{3}{2}} \\
\mathrm{~T}_2 & =8.49 \text { (or) } 6 \sqrt{2} \text { hours }
\end{aligned}
$

Question 2.
The time period of a satellite of Earth is 5 hours. If the separation between the Earth and the satellite is increased to four times the previous value, the new time period will become.
Answer:
$
\begin{aligned}
& \mathrm{T}_2=\mathrm{T}_1\left(\frac{a_2}{a_1}\right)^{\frac{3}{2}}=\mathrm{T}_1\left[\frac{4 a_1}{a_1}\right]^{\frac{3}{2}}=5(4)^{\frac{3}{2}}=5(8) \\
& \mathrm{T}_2=40 \text { hours }
\end{aligned}
$
Question 3.
The figure shows elliptical orbit of a planet ' $M$ ' about the Sun ' $S$ ', the shaded area SCD is twice the shaded area $\mathrm{SAB}$. If $\mathrm{t}_1$ is the time for the planet to move from $\mathrm{C}$ and $\mathrm{D}$ and $t_2$ is the time to move from $A$ to $B$ then.
Answer:

According to Kepler's law,
$
\begin{aligned}
\frac{\Delta \mathrm{A}}{\Delta t} & =\text { constant } \\
\frac{\Delta \mathrm{A}}{} & \propto \Delta t \\
\frac{\Delta \mathrm{A}_{\mathrm{SCD}}}{\Delta \mathrm{A}_{\mathrm{SAB}}} & =\frac{t_1}{t_2} \\
\frac{2 \mathrm{~A}}{\mathrm{~A}} & =\frac{t_1}{t_2} \\
t_1 & =2 t_2
\end{aligned}
$
Law of areas,
Question 4.
A satellite moves in a circle around the Earth, the radius of this circle is equal to one half of the radius of the Moon's orbit. The satellite completes one revolution in...... lunar month.
Answer:

Time period of revolution of moon around the Earth $\mathrm{T}_{\mathrm{m}}=1$ lunar month
$
\begin{aligned}
\frac{\mathrm{T}_e}{\mathrm{~T}_m} & =\left(\frac{a_e}{a_m}\right)^{\frac{3}{2}} \\
\mathrm{~T}_e & =\mathrm{T}_m \times\left(\frac{1}{2}\right)^{\frac{3}{2}}=1 \times(2)^{\frac{-3}{2}} \\
\mathrm{~T}_e & =2^{\frac{-3}{2}} \text { lunar month }
\end{aligned}
$
Question 5.
Two identical solid copper spheres of radius $\mathrm{R}$ are placed in contact with each other. The gravitational force between them is proportional to.
Answer:
The gravitational force $F=\frac{G(M)(M)}{(2 R)^2}=\frac{G\left(\frac{4}{3} \pi R^3 \rho\right)^2}{4 R^2}$.
Mass of the solid sphere $M=\frac{4}{3} \pi R^3 \rho$

$
\begin{array}{r}
F=\frac{G\left(16 \pi^2 R^6 \rho^2\right.}{9 \times 4 \times R^2} \\
F=\frac{4}{9} G \pi^2 \rho^2 R^4 ; F \propto \frac{4}{9} \pi^2 \rho^2 R^4
\end{array}
$
$\therefore$ The gravitational force between them is proportional to 4 th power of radius.
Question 6.
Two satellites A and B of the same mass are revolving around the Earth in circular orbits such that the distance of $B$ from the centre of the Earth is thrice as compared to the distance of A from the centre. What will be the ratio of centripetal force on B to that on $\mathrm{A}$.
Answer:
The necessary centripetal force is provided by the gravitational force of attraction, for circular orbit

Force on $\mathrm{A}, \frac{m v^2}{r}=\frac{\mathrm{GM} m}{r^2}$
Force on B, $\frac{m v^2}{r}=\frac{\mathrm{GM} m}{(3 r)^2}$
The ratio of centripetal force,
$
\begin{aligned}
& \frac{\mathrm{F}_{\mathrm{B}}}{\mathrm{F}_{\mathrm{A}}}=\frac{\mathrm{GM} m}{9 r^2} \times \frac{r^2}{\mathrm{GMm}} \\
& \frac{\mathrm{F}_{\mathrm{B}}}{\mathrm{F}_{\mathrm{A}}}=\frac{1}{9}
\end{aligned}
$
Question 7.
An infinite number of bodies, each of mass $2 \mathrm{~kg}$, are situated on $\mathrm{x}$-axis at distances $\mathrm{lm}$, $2 \mathrm{~m}, 4 \mathrm{~m}, 8 \mathrm{~m}$ from the origin. What will be the resultant gravitational potential due to this system at the origin.
Answer:
Gravitational potential at the origin is
$
\mathrm{V}=-\mathrm{G}\left[\frac{m}{r_1}+\frac{m}{r_2}+\frac{m}{r_3}+\frac{m}{r_4}+\ldots\right]=-\mathrm{G}\left[\frac{2}{1}+\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\ldots .\right]
$

$
=-2 G\left[1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots .\right]=-2 G\left[\frac{1}{1-\left(\frac{1}{2}\right)}\right]
$
$
\mathrm{V}=-4 \mathrm{G}
$
Question 8.
A body of mass ' $\mathrm{m}$ ' $\mathrm{kg}$ starts falling from a point $2 \mathrm{R}$ above the Earth's surface. What is its K.E. When it has fallen to a point ' $R$ ' above the Earth's surface.

Answer:
Potential energy
$
\begin{aligned}
\mathrm{U} & =\frac{-\mathrm{GMm}}{r}=\frac{\mathrm{GMm}}{\mathrm{R}+h} \\
\mathrm{U}_{\text {initial }} & =-\frac{\mathrm{GMm}}{r} \text { and } \mathrm{U}_{\text {final }}=-\frac{\mathrm{GMm}}{2 \mathrm{R}}
\end{aligned}
$
Loss of P.E. = gain in K.E. $=\frac{\mathrm{GMm}}{2 \mathrm{R}}-\frac{\mathrm{GMm}}{3 \mathrm{R}}=\frac{\mathrm{GM} m}{6 \mathrm{R}}$
$
\therefore \quad \mathrm{K} . \mathrm{E} .=\frac{\mathrm{GMm}}{6 \mathrm{R}}
$

Question 9.
A particle of mass $10 \mathrm{~g}$ is kept on the surface of a uniform sphere of mass $100 \mathrm{~kg}$ and radius $10 \mathrm{~cm}$. Find the work to done against the gravitational force between them to take the particle is away from the sphere.
Answer:
$
\begin{aligned}
& \mathrm{U}=\frac{-\mathrm{GMm}}{\mathrm{R}}=\frac{-6.67 \times 10^{-11} \times 100 \times 10 \times 10^{-3}}{10 \times 10^{-2}} \\
& \mathrm{U}=6.67 \times 10^{-10} \mathrm{~J}
\end{aligned}
$
So, the amount of work done to take the particle upto infinite will be $6.67 \times 10^{-10} \mathrm{~J}$
Question 10.
The mass of a space ship is $1000 \mathrm{~kg}$. It is to be launched from Earth 's surface out into free space the value of $g$ and $\mathrm{R}$ (radius of Earth) are $10 \mathrm{~ms}^{-2}$ and $6400 \mathrm{~km}$ respectively. The required energy for this work will be.
Answer:
Potential energy $\mathrm{U}=-\mathrm{mgR}_{\mathrm{e}}+\mathrm{mgh}$
(The first term is independent of the height, so it can be taken to zero.)
$
\begin{aligned}
& \mathrm{W}=\mathrm{U}=\mathrm{mgh}[\mathrm{h} \approx \mathrm{R}] \\
& =1000 \times 10 \times 6400 \times 10^3=64 \times 10^9 \\
& \mathrm{~W}=6.4 \times 10^{10} \mathrm{~J}
\end{aligned}
$
Question 11.
If the mean radius of the Earth is $\mathrm{R}$, its angular velocity is $\omega$, and the acceleration due to gravity at the surface of the Earth is $g$, then what will be the cube of the radius of the orbit of a geostationary satellite.
Answer:

Let $r$ be the radius of the geostationary orbit. Angular velocity of revolution of a
geostationary satellite is same as the angular velocity of rotation of the Earth.
$
\begin{aligned}
m r \omega^2 & =\frac{\mathrm{GM} m}{r^2} \\
r^3 & =\frac{\mathrm{GM} m}{m \omega^2} \times \frac{\mathrm{R}^2}{\mathrm{R}^2}=\frac{\mathrm{GM}}{\mathrm{R}^2}\left(\frac{\mathrm{R}^2}{\omega^2}\right) \quad\left[g=\frac{\mathrm{GM}}{\mathrm{R}^2}\right] \\
r^3 & =\frac{g \mathrm{R}^2}{\omega^2}
\end{aligned}
$
Question 12.
The escape velocity of a body from Earth's surface is $\mathrm{v}_{\mathrm{e}}$. What will be the escape velocity of the same body from a height equal to 7R from Earth's surface.
Answer:

$v_e \propto \frac{1}{\sqrt{r}}$ where $r$ is the position of body from the surface.
$
\begin{aligned}
\frac{v_e}{v^{\prime}}= & \sqrt{\frac{r^{\prime}}{r}}=\sqrt{\frac{\mathrm{R}+7 \mathrm{R}}{\mathrm{R}}}=\sqrt{\frac{8 \mathrm{R}}{\mathrm{R}}}=\sqrt{8}=2 \sqrt{2} \\
& \therefore v^{\prime}=\frac{v_e}{2 \sqrt{2}}
\end{aligned}
$
Question 13.
If $R$ is the radius of the Earth and $g$ is the acceleration due to gravity on the Earth's surface. Find the mean density of the Earth.

Acceleration due to gravity $g=\frac{G M}{R^2}$
Acceleration due to gravity $g=\frac{G M}{R^2}$
Mass of the Earth $\mathrm{M}=\frac{4}{3} \pi R^3 \rho$
$
\begin{array}{r}
\quad g=\frac{G}{\mathrm{R}^2}\left[\frac{4}{3} \pi \mathrm{R}^3 \rho\right] \\
\rho=\frac{3 g}{4 \pi \mathrm{RG}}
\end{array}
$
Question 14.
If the mass of Earth is 80 times of that of a planet and diameter is double that of planet and ' $\mathrm{g}$ ' on the Earth is $9.8 \mathrm{~ms}-2$. Calculate the value of ' $\mathrm{g}$ ' on that planet?
Answer:
Acceleration due to gravity $g=\frac{\mathrm{GM}}{\mathrm{R}^2}$
$
\begin{aligned}
& g_p=\frac{\mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^2} \text { and } g_e=\frac{\mathrm{GM}_e}{\mathrm{R}_e^2} \\
& \frac{g_p}{g_e}=\frac{\mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^2} \times \frac{\mathrm{R}_e^2}{\mathrm{GM}_e}=\frac{\mathrm{M}_{\mathrm{P}}}{\mathrm{M}_e}\left(\frac{\mathrm{R}_e}{\mathrm{R}_{\mathrm{P}}}\right)^2
\end{aligned}
$

Question 15.
At what distance from the centre of Earth, the value acceleration due to gravity ' $\mathrm{g}$ ' will be half that of the surface?
Answer:
According to acceleration due to gravity
$
\begin{aligned}
\frac{g^{\prime}}{g} & =\left(\frac{\mathrm{R}}{\mathrm{R}+h}\right)^2 \\
\frac{1}{2} & =\left(\frac{\mathrm{R}}{\mathrm{R}+h}\right)^2 \Rightarrow \frac{1}{\sqrt{2}}=\frac{\mathrm{R}}{\mathrm{R}+h} \\
\mathrm{R}+h & =\sqrt{2} \mathrm{R} \Rightarrow h=(\sqrt{2}-1) \mathrm{R}
\end{aligned}
$

$
h=0.414 \mathrm{R}
$
Hence distance from centre $=\mathrm{R}+0.414 \mathrm{R}=1.414 \mathrm{R}$
Question 16.
A body weight $700 \mathrm{~g}$ on the surface of Earth. How much it weight on the surface of planet whose mass is $\frac{1}{7}$ and radius is half that of the Earth.
Answer:
Acceleration due to gravity $g=\frac{G M}{R^2}$
$
\mathrm{M}_{\mathrm{e}}=\frac{\mathrm{M}}{7} \text { and } \mathrm{R}_e=\frac{\mathrm{R}}{2}
$
On the planet $g_p=\frac{\mathrm{G}\left(\frac{\mathrm{M}}{7}\right)}{\left(\frac{\mathrm{R}}{2}\right)^2}=\frac{4}{7}\left(\frac{\mathrm{GM}}{\mathrm{R}^2}\right)=\frac{4}{7} g$
Hence weight on the planet $W_p=700 \times \frac{4}{7}$
$
\mathrm{W}_{\mathrm{p}}=400 \text { gram }
$

Question 17.
An object weight $72 \mathrm{~N}$ on the Earth. What its weight at a height $\frac{\mathbf{R}}{2}$ from Earth.
Answer:
We know that
$
\begin{aligned}
& g=\frac{\mathrm{GM}}{} \\
& g^{\prime}=\mathrm{g}\left(\frac{\mathrm{R}}{\mathrm{R}+h}\right)^2=g\left(\frac{\mathrm{R}}{\mathrm{R}+\frac{\mathrm{R}}{2}}\right)^2=g\left(\frac{2 \mathrm{R}}{3 \mathrm{R}}\right)^2 \\
& g^{\prime}=\frac{4}{9} g
\end{aligned}
$

$
\text { Weight } \begin{aligned}
\mathrm{W}^{\prime} & =\frac{4}{9} \mathrm{~W}=\frac{4}{9}(72) \\
\mathrm{W}^{\prime} & =32 \mathrm{~N}
\end{aligned}
$
Question 18.
A body weight $500 \mathrm{~N}$ on the surface of the Earth. How much would it weight half way below the surface of Earth.
Answer:
Weight on surface of Earth, $\mathrm{mg}=500 \mathrm{~N}$
Weight below the surface of Earth at $\mathrm{d}=\frac{\mathrm{R}}{2}$
From variation of ' $g$ ' with depth
$
\begin{aligned}
\mathrm{g}^{\prime} & =g\left(1-\frac{d}{\mathrm{R}}\right) \\
m g^{\prime} & =m g\left[1-\frac{\left(\frac{\mathrm{R}}{2}\right)}{\mathrm{R}}\right]=500\left[1-\frac{1}{2}\right]=500\left[\frac{1}{2}\right] \\
\mathrm{W}^{\prime} & =250 \mathrm{~N}
\end{aligned}
$