Question

The same 1710 kg artificial satellite is placed into circular orbit at the same altitude of 2.6x10° m around an exoplanet with the same radius as the Earth, but twice the mass. a. What is the orbital speed of the satellite? b. What is the period of the satellite? C. What is the kinetic energy of the satellite? d. What is the total energy of the satellite?

Answer 1

Given:

mass of satellite, m = 1710 kg

altitude, h = $2.6\times 10^{6} m$

G =  $6.67\times 10^{-11}$

we know

mass of earth, $M_{E}$ =  $5.972\times 10^{24} kg$

Here, according to question we will consider

$2M_{E}$ =  $11.944\times 10^{24} kg$

radius of earth,  $R_{E}$ =  $6.371\times 10^{6} m$

Formulae Used and replacing $M_{E}$ by  $2M_{E}$ :

1). $v = \sqrt{\frac{2GM_{E}}{R_{E} + h}}$

2). $T = \sqrt{\frac{4\pi ^{2}(R_{E} + h)^{3}}{2GM_{E}}}$

3). $KE = \frac{1}{2}mv^{2}$

4). $Total Energy, E = -\frac{2GM_E\times m}{2(R_{E} + h)}$

where,

v = orbital velocity of satellite

T = time period

KE = kinetic energy

Solution:

Now, Using Formula (1), for orbital velocity:

 $v = \sqrt{\frac{6.67 \times 10^{-11} \times 11.944 \times 10^{24}}{6.371 \times 10^{6} + 2.6 \times 10^{6}}$

v =  $9.423 \times 10^{3}$  m/s

Using Formula (2) for time period:

$T = \sqrt{\frac{4\pi ^{2}(6.371\times 10^{6} + 2\times 10^{6})^{3}}{6.67\times 10^{-11}\times 9.44\times 10^{24}}}$

$T = 6.728\times 10^{3} s$

Now, Using Formula(3) for kinetic energy:

$KE = \frac{1}{2}(9.44\times 10^{24})(9.42\times 10^{3})^{2}$

$KE = \frac{1}{2}(1710)(9.42\times 10^{3})^{2} = 7.586\times 10^{10} J$

Now, Using Formula(4) for Total energy:

$E = -\frac{6.67\times 10^{-11}\times 9.44\times 10^{24}\times 1710}{2( 6.371\times 10^{6} + 2.6\times 10^{6})}$

$E = - 7.59\times 10^{10} J$