Numerical Problems-1 - Chapter 6 Gravitation 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
An unknown a planet orbits the Sun with distance twice the semi major axis distance of the Earth's orbit. If the Earth's time period is $\mathrm{Tp}$ what is the time period of this
unknown planet?
Answer:
By Kepler's 3rd law $\mathrm{T}^2 \propto \mathrm{a}^3$
Time period of unknown planet $=T_2$
Time period of Earth $=\mathrm{T}_1$
Distance of unknown planet from the Sun $=a_2$
Distance of the Earth from the Sun $=a_1$
$
\begin{aligned}
\frac{\mathrm{T}_1^2}{\mathrm{~T}_2^2} & =\frac{a_1^3}{a_2^3} \\
\mathrm{~T}_2 & =\left(\frac{a_2}{a_1}\right)^{\frac{3}{2}} \mathrm{~T}_1 \quad a_2=2 a, \\
\mathrm{~T}_2 & =\left(\frac{2 a_1}{a_1}\right)^{\frac{3}{2}} \mathrm{~T}_1 \Rightarrow \mathrm{T}_2=2 \sqrt{2} \mathrm{~T}_1
\end{aligned}
$
Question 2.
Assume that you are in another solar system and provided with the set of data given below consisting of the planets' semi major axes and time periods. Can you infer the relation connecting semi major axis and time period?
Answer:

In a given datas tells us the relation connecting to the semi major axis is proportional to the two times of square of the time period.
$a \propto 2 \mathrm{~T}^2$
Question 3.
If the masses and mutual distance between the two objects are doubled, what is the change in the gravitational force between them?
Answer:
By Newton's law of gravitation
$
\mathrm{F}=\frac{\mathrm{GM}_1 \mathrm{M}_2}{r^2}
$
Here, the masses and mutual distance between the two objects are doubled .
$
\mathrm{F}=\frac{\mathrm{G}\left(\mathrm{M}_1\right)\left(2 \mathrm{M}_2\right)}{(2 r)^2}=\frac{4 \mathrm{GM}_1 \mathrm{M}_2}{4 r^2}
$
$
\mathrm{F}=\frac{\mathrm{GM}_1 \mathrm{M}_2}{r^2}
$
There is no change in the gravitational force between them.
Question 4.
Two bodies of masses $\mathrm{m}$ and $4 \mathrm{~m}$ are placed at a distance $r$. Calculate the gravitational potential at a point on the line joining them where the gravitational field is zero.
Answer:
Let the point be the position when the gravitational field is zero.,
$
\frac{\mathrm{G} m}{x^2}=\frac{\mathrm{G}(4 m)}{(r-x)^2}
$

$
\frac{1}{x^2}=\frac{4}{(r-x)^2}
$
Square root on both sides, $\frac{1}{x}=\frac{2}{(r-x)}$
$
\therefore \quad x=\frac{r}{3}
$
The point $\mathrm{P}$ is at a distance $\frac{r}{3}$ from mass ' $\mathrm{m}$ ' and $\frac{2 r}{3}$ from mass ' $4 \mathrm{~m}$ '
Gravitational potential
$
\begin{aligned}
\mathrm{V} & =\frac{-\mathrm{Gm}}{\left(\frac{r}{3}\right)}-\frac{\mathrm{G}(4 m)}{\left(\frac{2 r}{3}\right)} \\
& =\frac{-3 G m}{4}-\frac{3 G(4 m)}{2 r}=\frac{-3 G m}{r}-\frac{12 G m}{2 r} \Rightarrow V=\frac{-9}{}
\end{aligned}
$

Question 5.
If the ratio of the orbital distance of two planets $\frac{d_1}{d_2}=2$, what is the ratio of gravitational field experienced by these two planets?
Answer:
The gravitational field experienced by planets 1
$
\mathrm{E}_1=\frac{\mathrm{GM}}{d_1^2}
$
The gravitational field experienced by planet 2
$
\mathrm{E}_2=\frac{\mathrm{GM}}{d_2^2}
$
The ratio of their orbital distance $\frac{d_1}{d_2}=2$
$
\therefore \quad d_1=2 d_2
$
The ratio of gravitational field of two planets.
$
\begin{aligned}
\frac{\mathrm{E}_1}{\mathrm{E}_2} & =\frac{\mathrm{GM}}{\left(2 d_2\right)^2} \times \frac{d_2^2}{\mathrm{GM}}=\frac{\mathrm{GM}}{4 d_2^2} \times \frac{d_2^2}{\mathrm{GM}} \\
\frac{\mathrm{E}_1}{\mathrm{E}_2} & =\frac{1}{4} \\
\therefore \quad \mathrm{E}_2 & =4 \mathrm{E}_1
\end{aligned}
$

Question 6.
The moon $\mathrm{I}_{\mathrm{o}}$ orbits Jupiter once in 1.769 days. The orbital radius of the Moon $\mathrm{I}_0$ is $421700 \mathrm{~km}$. Calculate the mass of Jupiter?
Answer:
Kepler's third law is used to find the mass of the planet

$
\begin{aligned}
\mathrm{T}^2 & =\frac{4 \pi^2}{\mathrm{GM}}(\mathrm{R}+h)^2 \\
\mathrm{M} & =\frac{4 \pi^2}{\mathrm{GT}^2}(\mathrm{R}+h)^2=\frac{4(3.14)^2 \times\left(421700 \times 10^3\right)^2}{6.67 \times 10^{-11} \times(1.769 \times 86400)^2} \\
& =\frac{39.4384 \times 7.499128631 \times 10^{25}}{6.67 \times 10^{-11} \times 2.336055469 \times 10^{10}} \\
& =\frac{295.7536 \times 10^{25}}{1.5581}=189.8169 \times 10^{25} \\
\mathrm{M} & =1.898 \times 10^{27} \mathrm{~kg}
\end{aligned}
$
Question 7.
If the angular momentum of a planet is given by $\overrightarrow{\mathbf{L}}=5 t^2 \hat{i}-6 \hat{t} \hat{j}+3 \hat{k}$. What is the torque experienced by the planet? Will the torque be in the same direction as that of the angular momentum?
Answer:
The torque experienced by the planet
$
\vec{\tau}=\frac{d \overrightarrow{\mathrm{L}}}{d t} \quad\left[\frac{d}{d t}\left(t^n\right)=n t^{n-1}\right]
$

$
\begin{aligned}
& =\frac{d}{d t}\left(5 t^2 \hat{i}-6 t \hat{j}+3 \hat{k}\right) \\
& =5 \frac{d}{d t}\left(t^2\right) \hat{i}-6 \frac{d}{d t}(t) \hat{j}+3 \frac{d}{d t}(\hat{k})=10 t \hat{i}-6 \hat{j}+0 \\
\vec{\tau} & =10 t \hat{i}-6 \hat{j}
\end{aligned}
$
Question 8 .
Four particles, each of mass $\mathrm{M}$ and equidistant from each other, move along a circle of radius $\mathrm{R}$ under the action of their mutual gravitational attraction. Calculate the speed of each particle.
Answer:
The net gravitational force $=$ Centripetal force

Force $\mathrm{F}_1$ between $\mathrm{A}$ and $\mathrm{B}, \quad \mathrm{F}_1=\frac{\mathrm{GMM}}{(\sqrt{2} R)^2}$
Force $F_2$ between $A$ and $D, \quad F_2=\frac{G M M}{(\sqrt{2} R)^2}$
Force $\mathrm{F}_3$ between $\mathrm{A}$ and $\mathrm{C}$ (or $\mathrm{B}$ and $\mathrm{D}$ )
$
\mathrm{F}_3=\frac{\mathrm{GMM}}{(2 \mathrm{R})^2}
$
The components $\mathrm{F} 1$ and $\mathrm{F} 2$ along the radius
$
\begin{aligned}
\mathrm{F}_1 \cos 45 \text { and } \mathrm{F}_2 \cos 45 \mathrm{~F}_1 & =\mathrm{F}_2=\mathrm{F} \\
\therefore \quad \text { Net force } & =2 \mathrm{~F} \cos 45+\mathrm{F}_3 \\
& =2 \frac{\mathrm{GM}^2}{(\sqrt{2} R)^2} \times\left(\frac{1}{2}\right)+\frac{\mathrm{GM}^2}{4 \mathrm{R}^2} \\
\frac{\mathrm{M} v_0^2}{\mathrm{R}} & =\frac{\mathrm{GM}^2}{2 \mathrm{R}^2}(2 \sqrt{2}+1)
\end{aligned}
$

$
\begin{aligned}
& v_0^2=\frac{\mathrm{GM}}{4 \mathrm{R}}[1+2 \sqrt{2}] \\
& v_o=\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}
\end{aligned}
$
Question 9.
Suppose unknowingly you wrote the universal gravitational constant value as $\mathrm{G}=6.67$ $\times 10^{11}$ instead of the correct value $\mathrm{G}=6.67 \times 10^{-11}$, what is the acceleration due to gravity g' for this incorrect G? According to this new acceleration due to gravity, what will be your weight $\mathrm{W}$ ?
Answer:
Data: Incorrect Gravitational constant $\mathrm{G}=6.67 \times 10^{11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}$
Mass of the Earth $\mathrm{M}_{\mathrm{e}}=5.972 \times 10^{24} \mathrm{~kg}$
Radius of the earth $\mathrm{R}_{\mathrm{e}}=6371 \mathrm{~km}$ (or) $6371 \times 10^3 \mathrm{~m}$
Acceleration due to gravity $g^{\prime}=\frac{\mathrm{GM}_e}{\mathrm{R}_e^2}$
$
\begin{aligned}
g^{\prime} & =\frac{6.67 \times 10^{11} \times 5.97 \times 10^{24}}{\left(6371 \times 10^3\right)^2} \\
& =\frac{39.8332 \times 10^{35}}{4.0589641 \times 10^{13}}=9.81 \times 10^{22}
\end{aligned}
$

New weight
$
\begin{aligned}
& \mathrm{W}^{\prime}=\mathrm{M} g^{\prime}=\mathrm{M}\left(10^{22} \mathrm{~g}\right)=10^{22} \mathrm{mg} \\
& \mathrm{W}^{\prime}=10^{22} \mathrm{~W}
\end{aligned}
$
Question 10.
Calculate, the gravitational field at point $\mathrm{O}$ due to three masses $\mathrm{m}_1, \mathrm{~m}_2$, and $\mathrm{m}_3$ whose positions are given by the following figure. If the masses $m_1$ and $m_2$ are equal what is the change in gravitational field at the point $\mathrm{O}$ ?
Answer:

Gravitational field due to ' $m_1$ ' at a point ' $\mathrm{O}$ ' is $\overrightarrow{\mathrm{E}}_1=\frac{\mathrm{G} m_1}{a^2} \hat{i}$
Gravitational field due to ' $m_1$ ' at a point ' $\mathrm{O}$ ' is $\overrightarrow{\mathrm{E}}_2=-\frac{\mathrm{G} m_2}{a^2} \hat{i}$
(Negative sign indicates field acting along negative $x$ direction)
Gravitational field due to ' $m_3$ ' at a point ' $\mathrm{O}$ ' is $\overrightarrow{\mathrm{E}}_3=\frac{\mathrm{G} m_3}{a^2} \hat{j}$
(Negative sign indicates field acting along negative $\mathrm{x}$ direction)

$
\begin{aligned}
& \overrightarrow{\mathrm{E}}=\overrightarrow{\mathrm{E}}_1+\overrightarrow{\mathrm{E}}_2+\overrightarrow{\mathrm{E}}_3=\frac{\mathrm{G} m_1}{a^2} \hat{i}-\frac{\mathrm{G} m_2}{a^2} \hat{i}+\frac{\mathrm{G} m_3}{a^2} \hat{j} \\
& \overrightarrow{\mathrm{E}}=\frac{\mathrm{G}}{a^2}\left[\left(m_1-m_2\right) \hat{i}+m_3 \hat{j}\right]
\end{aligned}
$
If the masses $m_1=m_2$, then
$
\overrightarrow{\mathrm{E}}=\frac{\mathrm{G} m_3}{a^2} \hat{j}
$
Question 11.
What is the gravitational potential energy of the Earth and Sun? The Earth to Sun distance is around 150 million $\mathrm{km}$. The mass of the Earth is $5.9 \times 10^{24} \mathrm{~kg}$ and mass of the Sun is $1.9 \times 10^{30} \mathrm{~kg}$.

Answer:
Mass of the Earth $\mathrm{M}_{\mathrm{E}}=5.9 \times 10^{24} \mathrm{~kg}$
Mass of the Sun $\mathrm{M}_{\mathrm{S}}=1.9 \times 10^{30} \mathrm{~kg}$
Distance between the Sun and Earth
$\mathrm{r}=150$ million $\mathrm{km} ; \mathrm{r}=150 \times 10^9 \mathrm{~m}$
Gravitational constant $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}$
The gravitational potential energy
$
\begin{aligned}
\mathrm{U} & =-\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{M}_{\mathrm{S}}}{r}=\frac{-6.67 \times 10^{-11} \times 5.9 \times 10^{24} \times 1.9 \times 10^{30}}{150 \times 10^9} \\
& =-\frac{74.7707}{150} \times 10^{34}=-0.4985 \times 10^{34} \\
U & =-4.985 \times 10^{33} \text { joule (or) } \mathrm{J}
\end{aligned}
$
Question 12.
Earth revolved around the Sun at $30 \mathrm{~km} \mathrm{~s}^{-1}$. Calculate the kinetic energy of the Earth. In the previous example you calculated the potential energy of the Earth. What is the total energy of the Earth in that case? Is the total energy positive? Give reasons.
Answer:
Mass of the Earth $\mathrm{M}_{\mathrm{E}}=5.9 \times 10^{24} \mathrm{~kg}$
Speed of the Earth rovolves around the Sun
$
\begin{aligned}
& \mathrm{V}_{\mathrm{E}}=30 \mathrm{k} \mathrm{ms}^{-1} \\
& \mathrm{~V}_{\mathrm{E}}=30 \times 10^3 \mathrm{~ms}^{-1}
\end{aligned}
$

Kinetic energy of Earth
$
\begin{aligned}
\mathrm{E}_{\mathrm{K}} & =\frac{1}{2} \mathrm{M}_{\mathrm{E}} \mathrm{V}_{\mathrm{E}}^2 \\
\mathrm{E}_{\mathrm{K}} & =\frac{1}{2} \times 5.9 \times 10^{24} \times\left(30 \times 10^3\right)^2 \\
& =\frac{1}{2} \times 5310 \times 10^{30}=2655 \times 10^{30} \\
\mathrm{E}_{\mathrm{K}} & =2.655 \times 10^{33} \text { joule (or) } \mathrm{J}
\end{aligned}
$
Total energy of the Earth, $\quad E=$ Kinetic energy + Potential energy
$
=\frac{1}{2} \mathrm{MV}^2+\left(-\frac{\mathrm{GM}_1 \mathrm{M}_2}{r^2}\right)
$

$
\begin{aligned}
& =2.655 \times 10^{33}+\left(-4.985 \times 10^{33}\right) \\
& =2.655 \times 10^{33}-4.985 \times 10^{33} \\
& E=-2.33 \times 10^{33} \text { joule (or) } \mathrm{J}
\end{aligned}
$
'-Ve' implies that Earth is bounded with Sun.
Question 13.
An object is thrown from Earth in such a way that it reaches a point at infinity with nonzero kinetic energy $\left[\mathbf{K} \cdot \mathbf{E}(r=\infty)=\frac{1}{2} \mathbf{M} v_{\infty}^2\right]$, with that velocity should the object be thrown from Earth?
Answer:
An object is thrown up with an initial velocity is $\mathrm{v}_{\mathrm{i}}$, So Total energy of the object is
$
\mathrm{E}_i=\frac{1}{2} \mathrm{M}_i^2-\frac{\mathrm{GMm}_e}{\mathrm{R}_{\mathrm{E}}}
$
Now, the object reaches a height with a non-zero K.E.
K.E becomes infinity. P.E becomes zero.
So,
$
\begin{aligned}
\mathrm{E}_f & =\frac{1}{2} \mathrm{M} v_{\infty} \\
\mathrm{E}_i & =\mathrm{E}_f \Rightarrow v_{\mathrm{i}}=\underline{\mathrm{V}}_{\infty} \\
\frac{1}{2} \mathrm{M} v_{\infty}^2 & =\frac{1}{2} \mathrm{M} v_e^2-\frac{2 \mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}
\end{aligned}
$

$
\begin{aligned}
\frac{1}{2} \mathrm{M}_{\infty}^2 & =\frac{1}{2} \mathrm{M}\left[\mathrm{V}_e^2-\frac{2 \mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\right] \\
v_e^2 & =v_{\infty}^2+2\left(\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\right) \mathrm{R}_{\mathrm{E}} \\
& =v_{\infty}^2+2 \mathrm{gR}_{\mathrm{E}} \\
v_e & =\sqrt{v_{\infty}^2+2 g \mathrm{R}_{\mathrm{E}}}
\end{aligned}
$
Question 14.
Suppose we go $200 \mathrm{~km}$ above and below the surface of the Earth, what are the g values
at these two points? In which case, is the value of $g$ small?
Answer:

Variation of g' with depth
$
\begin{aligned}
g^{\prime} & =g\left(1-\frac{d}{\mathrm{R}_{\mathrm{E}}}\right)\left[\begin{array}{c}
d=200 \mathrm{~km}=200 \times 10^3 \mathrm{~m} \\
\mathrm{R}_{\mathrm{E}}=6371 \times 10^3 \mathrm{~m}
\end{array}\right] \\
& =g\left(1-\frac{200 \times 10^3}{6371 \times 10^3}\right)=g(1-0.0314)=g(0.9686) \\
g^{\prime} & =0.96 g
\end{aligned}
$
Variation of g' with altitude
$
\begin{aligned}
g^{\prime} & =g\left(1-\frac{2 h}{\mathrm{R}_{\mathrm{E}}}\right) \quad\left[h=200 \mathrm{~km}=200 \times 10^3 \mathrm{~m}\right] \\
& =g\left(1-\frac{2 \times 200 \times 10^3}{6371 \times 10^3}\right)=g(1-2(0.0314))=g(0.9372) \\
g^{\prime} & =0.93 \mathrm{~g}
\end{aligned}
$
Question 15.
Calculate the change in $g$ value in your district of Tamil Nadu. (Hint: Get the latitude of your district of Tamilnadu from the Google). What is the difference in g values at " Chennai and Kanyakumari?
Answer:
Variation of ' $\mathrm{g}$ ' value in the latitude to chennai

$
g_{\text {Chennai }}^{\prime}=g-\omega^2 \mathrm{R} \cos ^2 \lambda
$
Here
$
\omega^2 R=\left(\frac{2 \pi}{T}\right)^2 \times R
$
Period of revolution $(\mathrm{T})=1$ day $=86400 \mathrm{sec}$
Radius of the Earth $(\mathrm{R})=6400 \times 10^3 \mathrm{~m}$
Latitude of Chennai $(\lambda)=13^{\circ}=0.2268 \mathrm{rad}$
$
g_{\text {Chennai }}^{\prime}=9.8-\left[\left(\frac{2 \times 3.14}{86400}\right)^2 \times 6400 \times 10^3\right] \times(\cos 0.2268)^2
$

$
\begin{aligned}
& =9.8-\left[\left(3.4 \times 10^{-2}\right) \times(0.9744)^2\right] \\
& =9.8-[0.034 \times 0.9494]=9.8-0.0323 \\
g_{\text {Chennai }}^{\prime} & =9.7677 \mathrm{~ms}^{-2}
\end{aligned}
$
Variation of ' $\mathrm{g}$ ' value in the latitude of Kanyakumari
$
\begin{aligned}
& \lambda_{\text {Kanyakumari }}^{\prime}=8^{\circ} 35^{\prime}=0.1457 \mathrm{rad} \\
& g_{\text {Kanyakumari }}^{\prime}=9.8-\left[3.4 \times 10^{-2} \times(\cos 0.1457)^2\right]=9.8-0.033 \mathrm{Z} \\
& g_{\text {Kanyakumari }}^{\prime}=9.7667 \mathrm{~ms}^{-2}
\end{aligned}
$
The difference of ' $g$ ' value $\Delta g=g_{\text {Chennai }}-g_{\text {Kanyamukari }}^{\prime}$
$
=9.7677-9.7667
$
- $\Delta g=0.001 \mathrm{~ms}^{-2}$