Question

Two satellites ${S_1}$ & ${S_2}$of equal masses revolves in the same sense around a heavy planet in a coplanar circular orbit of radii $R$ and $4R$:
$\left( a \right)$ The ratio of the period of revolution ${S_1}$ & ${S_2}$ is $1:8$
$\left( b \right)$ Their velocities are in the ratio $2:1$
$\left( c \right)$ Their angular momentum about the planet are in the ratio $2:1$
$\left( d \right)$ The ratio of angular velocities of ${S_1}$ & ${S_2}$ w.r.t. ${S_1}$ when all three are in the same line is$9:5$.

Answer 1

So for solving this question there are a total of four terms used here. The first one is the period of the revolution another one is their velocities and the angular momentum. So the last one is the ratio of their angular velocities.

Formula used:
Period,
$\Rightarrow T = 2\pi \sqrt {\dfrac{{{R^3}}}{{GM}}}$
Where $T$ is the period of the satellite, $R$will be the average radius of the orbit and$G = 6.673 \times {10^{ - 11}}N{m^2}/k{g^2}$.
Velocity,
$\Rightarrow V = \sqrt {\dfrac{{GM}}{R}}$
Also,
$\Rightarrow V = R\omega$
Where; $\omega$ is the angular velocity.
Angular momentum,
$\Rightarrow L = MVR$
Where, $L$ is the angular momentum, $V$ is the velocity, and $M$ is the mass.

Complete step by step solution:
$\left( a \right)$ We have to calculate and check the ratio of the period so we will do it by comparing both the period ratio.
$\Rightarrow T = 2\pi \sqrt {\dfrac{{{R^3}}}{{GM}}}$
Since the period is directly proportional to the radius, therefore
$\Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = {\left( {\dfrac{1}{4}} \right)^{3/2}}$
On solving the above
$\Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{1}{8}$
Hence the statement first is correct.
$\left( b \right)$. Now we have to check the velocity, so as we do the comparison of velocities, we get
$\Rightarrow V = \sqrt {\dfrac{{GM}}{R}}$
Since velocity is inversely proportional to the radius, therefore
$\Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = {\left( {\dfrac{4}{1}} \right)^{1/2}}$
On solving the above ratio, we get
$\Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{2}{1}$
Therefore, the second statement is also true.
$\left( c \right)$. Now we have to compare the angular velocity and since we have already calculated the value of velocity, we will put the value and solve the equation.
$\Rightarrow \dfrac{{{L_1}}}{{{L_2}}} = \dfrac{{M{V_1}R}}{{M{V_2}\left( {4R} \right)}}$
After putting the values,
$\Rightarrow \dfrac{{{L_1}}}{{{L_2}}} = \dfrac{2}{1} \times \dfrac{1}{4}$
We get,
$\Rightarrow \dfrac{{{L_1}}}{{{L_2}}} = \dfrac{1}{2}$
Therefore the third statement is wrong.

Therefore the right choice is: $A,B,D$

Note: Angular velocity could be a vector amount and is delineated because the rate of change of angular displacement that specifies the angular speed or rotational speed of an object and therefore the axis regarding that the object is rotating. The number of amendments of angular displacement of the particle at a given amount of your time is named angular velocity. The track of the angular velocity vector is vertical to the plane of rotation, during a direction that is typically indicated by the right-hand rule.