Question

Solve the following problems.
a. If mass of a planet is eight times the mass of the earth and its radius is twice the radius of the earth, what will be the escape velocity for that planet?
b. How much time a satellite in an orbit at height 35780 km above earth's surface would take, if the mass of the earth would have been four times its original mass?
c. If the height of a satellite completing one revolution around the earth in T seconds is h₁ meter, then what would be the height of a satellite taking $2\sqrt{2}\mathrm{T}$ seconds for one revolution?

Answer 1

a. Escape velocity for Earth is
$v_{\mathrm{esc}}=\sqrt{\frac{2G{M}_{\mathrm{e}}}{R_{\mathrm{e}}}}=11.2\mathrm{km}/\mathrm{s}$
Given:
Mass of the planet, M_p = 8$\times$Mass of the Earth (M_e)
Radius of the planet, R_p = 2$\times$Radius of the Earth (R_e)
Thus, escape velocity of the planet is
$v_{\mathrm{esc}}^{\text{'}}=\sqrt{\frac{2G{M}_{\mathrm{p}}}{R_{\mathrm{p}}}}=\sqrt{\frac{8}{2}}\sqrt{\frac{2G{M}_{\mathrm{e}}}{R_{\mathrm{e}}}}=\sqrt{4}\times 11.2=22.4\mathrm{km}/\mathrm{s}$

b. Given:
Height of the satellite, h = 35780 km
Let the original mass of Earth be M. Then its new mass will be 4M.
Time taken by the satellite to revolved around the Earth's orbit is given as
$T=\frac{2\mathrm{\pi R}}{v_c}$
Now, v_c is given as
$v_{\mathrm{c}}=\sqrt{\frac{GM}{R+h}}$
Thus,
$$\begin{gathered} T=\frac{2\pi (R+h)}{\sqrt{\frac{GM}{R+h}}}=\frac{2\pi (R+h)\sqrt{R+h}}{\sqrt{GM}}.....\left(\mathrm{i}\right) \\ \Rightarrow T\propto \frac{1}{\sqrt{M}}.....\left(\mathrm{ii}\right) \end{gathered}$$
Thus from equation (ii) we see that when the mass of the Earth becomes 4 times, the time period of revolution of satellite should be halved.
i.e. $T_{4M}=\frac{T}{2}$ .....(iii)
Now, h = 35780 km
M = $6\times {10}^{24}\mathrm{kg}$
R = $6.4\times {10}^5\mathrm{m}$
G = 6.67 × 10^-11 N m² /kg²
Putting the values of h, M and R in first, we get
T = 24 h
Using (iii), we get
$T_{4M}=\frac{24}{2}=12\mathrm{h}$

c. Time period of the satellite is given as
$T=\frac{2\pi (R+h)\sqrt{R+h}}{\sqrt{GM}}$
When the height of the satellite is h₁, it takes T time to revolve around the Earth. Thus,
$T=\frac{2\pi (R+h_1)\sqrt{R+h_1}}{\sqrt{GM}}=\frac{2\pi (R+h_1{)}^{{\displaystyle \frac{3}{2}}}}{\sqrt{GM}}$ .....(i)
When the satellite takes $2\sqrt{2}\mathrm{T}$ time to revolve around the Earth, let it be at height h₂. Thus,
$2\sqrt{2}T=\frac{2\pi (R+h_2)\sqrt{R+h_2}}{\sqrt{GM}}=\frac{2\pi {\left(R+h_2\right)}^{\frac{3}{2}}}{\sqrt{GM}}$ .....(ii)
Dividing (ii) by (i), we get
$$\begin{gathered} 2\sqrt{2}=\frac{{\left(R+h_2\right)}^{{\displaystyle \frac{3}{2}}}}{{\left(R+h_1\right)}^{\frac{3}{2}}} \\ \Rightarrow \frac{R+h_2}{R+h_1}=2 \\ \Rightarrow h_2=R+2h_1 \end{gathered}$$