3 Marks Question

Question 13 Marks

The radius of a planet is three times the radius of the Earth but the density of both is same. If $v_p$ on $v_e$ are the escape velocities on the planet and the earth then prove that $v_P=3 v_e$.

Answer

We know that escape velocity on Earth,
$v_e=\sqrt{\left(\frac{2 GM_e}{R_c}\right)}=\sqrt{\frac{2 G\left(4 / 3 \pi R_e^3 d_c\right)}{R_c}}$
Here $d_e$ is the mean density of earth,
$v_e=R_c \sqrt{\frac{8 \pi G d_c}{3}}.....(1)$
Similarly, escape velocity on planet,
$v_p=R_p \sqrt{\frac{8 \pi G d_p}{3}}.....(2)$
From equations (1) and (2)
$\frac{v_p}{v_c}=\frac{R_p}{R_c} \sqrt{\frac{d_p}{d_c}}$
According to question,
$d_p=d_e \text { and } R_p=3 R_e$
Put the value,
Hence,
$\frac{v_p}{v_c}=\frac{3 R_c}{R_c} \times \sqrt{\frac{d_c}{d_c}}=3$
Hence,
$v_p=3 v_e\quad \text { Ans. }$

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Question 23 Marks

Three objects of equal mass $M$ are located at the vertices of an equilateral triangle of side $a$. At what speed should the three bodies be rotated on a circle so that the triangle moves on the circumference of the circular chamber and the side of the triangle remains unchanged.

Answer

The distance of each object of mass M placed on the vertices of triangle $A B C$ from the center of the triangle will be $=\frac{a}{\sqrt{3}}$

Suppose all three bodies are rotated on a circle with speed $v$ so that the triangle moves on the circumference of the circular orbit then the force acting on each mass
$=\frac{Mv^2}{\frac{a}{\sqrt{3}}}$
The resultatnt force of attraction on each mass due to the other two masses
$\begin{array}{l}
=\frac{GM^2}{a^2} \cos 30^{\circ}+\frac{GM^2}{a^2} \cos 30^{\circ} \\
=\frac{GM^2}{a^2} \times \frac{\sqrt{3}}{2}+\frac{GM^2}{a^2} \times \frac{\sqrt{3}}{2} \\
=\frac{2 GM^2}{a^2} \times \frac{\sqrt{3}}{2}=\sqrt{3} \frac{GM^2}{a^2}
\end{array}$
The resultant force of gravity acting on each mass will be equal to the resultant centripetal force acting on it.
$\begin{aligned}
\therefore \quad \frac{Mv^2}{\frac{a}{\sqrt{3}}} & =\frac{\sqrt{3} GM^2}{a^2} \\
v^2 & =\frac{\sqrt{3} GM^2}{a^2} \times \frac{a}{\sqrt{3} \times M} \\
v^2 & =\frac{GM}{a} \\
v & =\sqrt{\frac{GM}{a}}
\end{aligned}$

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Physics 3 Marks Question

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