Question

A $200kg$ satellite is revolving around the earth in a circular orbit of radius $2r$. The amount of energy required to transfer it to a circular orbit of radius $4r$ is $[m_e=6x{10}^{24}kg,r=6.4\times {10}^{6}m]$

Answer 1

Step 1: Given Data

Mass of the satellite $m_s=200kg$

Initial radius of the orbit of the satellite $r_1=2r$

The final radius of the orbit of the satellite $r_2=4r$

Mass of the earth $m_e=6\times {10}^{24}kg$

The radius of the earth $r=6.4\times {10}^{6}m$

Step 2: Change in Energy Expression

We know that the energy of an orbiting body is given as,

$E=-\frac{G{m}_1{m}_2}{2R}$

where $G$ is the gravitational constant with a value of $6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$.

Upon substituting the variables we get the initial energy of the satellite as,

$E_i=-\frac{G{m}_e{m}_s}{4r}$ $\left(1\right)$

Similarly, the final energy of the satellite as,

$E_f=-\frac{G{m}_e{m}_s}{8r}$ $\left(2\right)$

Therefore, the change in energy is given as

$∆E=E_f-E_i$

$=-\frac{G{m}_e{m}_s}{8r}+\frac{G{m}_e{m}_s}{4r}$

$=\frac{-G{m}_e{m}_s+2G{m}_e{m}_s}{8r}$

$\therefore ∆E=\frac{G{m}_e{m}_s}{8r}$ $\left(3\right)$

Step 3: Calculate the Change in energy

We know that the acceleration due to gravity is given as

$g=\frac{G{m}_e}{r^2}$

$\Rightarrow G{m}_e=g{r}^2$

Upon substituting this in equation $\left(3\right)$ we get,

$∆E=\frac{g{r}^2{m}_s}{8r}$

$\Rightarrow ∆E=\frac{gr{m}_s}{8}$

We know that acceleration due to gravity $g=9.8m/s^2$.

Upon substituting the values we get

$∆E=\frac{9.8\times 6.4\times {10}^6\times 200}{8}$

$=1.56\times {10}^{9}J$

Hence, the amount of energy required to transfer it to the new circular orbit is $1.56\times {10}^{9}J$.