91Ó°ÊÓ

The period of the satellite is $T=8.00×{10}^{4}s$

The mass of the planet is $M=7.00×{10}^{24}kg$

The radius of elliptical orbit at aphelion is $R_a=4.5×{10}^{7}m$

The angular speed of the satellite at aphelion is${\mathrm{Ó\neg }}_a=7.158×{10}^{-5}\frac{\text{rad}}{\text{s}}$

The square of a planet's period of revolution around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis, states Kepler's law of periods.

A spinning system's conservation of angular momentum ensures that its spin stays constant unless it is perturbed by an outside force.

We can find the angular speed of a satellite at perihelion by using the law of conservation of angular momentum. Then putting the expression for the semi major axis in it from Kepler’s law of period, we can find the value of angular speed of a satellite at perihelion.

Formula:

$$\begin{gathered} I_p{\mathrm{Ó\neg }}_p=I_a{\mathrm{Ó\neg }}_a \\ {T}^2=\left(\frac{4{\mathrm{Ï€}}^2}{GM}\right)a^3 \end{gathered}$$

Where, G is the gravitational constant ( $6.67×{10}^{-11} \frac{N.m^2}{k{g}^2}$), I_p is moment of inertia at perihelion, ${\mathrm{Ó\neg }}_p$ is angular speed at perihelion, $I_a$ is moment of inertia at aphelion, ${\mathrm{Ó\neg }}_a$ is angular speed at aphelion

According to the law of conservation of angular momentum,

$$\begin{gathered} I_p{\mathrm{Ó\neg }}_p=I_a{\mathrm{Ó\neg }}_a \\ {\mathrm{Ó\neg }}_p=\frac{I_a}{I_p}{\mathrm{Ó\neg }}_a \end{gathered}$$

…(¾±)

According to the definition of the moment of inertia,

$$\begin{gathered} I_p=M{R}_p^2 \\ {I}_a=M{R}_a^2 \end{gathered}$$

Equation (i) becomes

$$\begin{gathered} {\mathrm{Ó\neg }}_p=\left(\frac{M{R}_a^2}{M{R}_p^2}\right){\mathrm{Ó\neg }}_a \\ {\mathrm{Ó\neg }}_p=\left(\frac{R_a^2}{R_p^2}\right){\mathrm{Ó\neg }}_a \end{gathered}$$

…(¾±¾±)

For the elliptical orbit,

$$\begin{gathered} R_a+R_p=2a \\ {R}_p=2a-R_a \end{gathered}$$

Equation (ii) becomes

${\mathrm{Ó\neg }}_p=\left[\frac{R_a^2}{{\left(2a-R_a\right)}^2}\right]{\mathrm{Ó\neg }}_a$ …(¾±¾±¾±)

According to Kepler’s period law, the square of the period of any planet is proportional to the cube of the semi-major axis of its orbit.

$$\begin{gathered} T^2=\left(\frac{4{\mathrm{Ï€}}^2}{GM}\right)a^3 \\ a={\left(\frac{GM{T}^2}{4{\mathrm{Ï€}}^2}\right)}^{\frac{1}{3}} \end{gathered}$$

Equation (iii) becomes

$$\begin{gathered} {\mathrm{Ó\neg }}_p=\left[\frac{R_a^2}{{\left(2{\left(\frac{GM{T}^2}{4{\mathrm{Ï€}}^2}\right)}^{\frac{1}{3}}-R_a\right)}^2}\right]{\mathrm{Ó\neg }}_a \\ =\left[\frac{{\left(4.5×{10}^7\text{m}\right)}^2}{{\left(2{\left(\frac{\left(6.67×{10}^{-11}\frac{{\text{Nm}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)×7.00×{10}^{24}\text{kg}×{\left(8.00×{10}^4\text{s}\right)}^2}{4{\left(3.142\right)}^2}\right)}^{\frac{1}{3}}-4.5×{10}^7\text{m}\right)}^2}\right]×\left(7.158×{10}^{-5}\text{rad/s}\right) \\ =9.24×{10}^{-5}\frac{\text{rad}}{\text{s}} \end{gathered}$$

Therefore, the angular speed of a satellite at perihelion is $9.24×{10}^{-5}\frac{rad}{s}$.