Question

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to orbit at an altitude of 7580 km. Satellite B is to orbit at an altitude of 22000 km. The radius of Earth REis 6370 km. (a) What is the ratio of the potential energy of satellite B to that of satellite A, in orbit? (b) What is the ratio of the kinetic energy of satellite B to that of satellite A, in orbit? (c) Which satellite (answer A or B) has the greater total energy if each has a mass of 18.2 kg? (d) By how much?

Answer 1

(a) 0.49

The potential energy of a satellite in orbit is given by

$U=-\frac{GMm}{R+h}$

where G is the gravitational constant, M is the Earth's mass, m is the satellite's mass, R is the Earth's radius and h is the altitude of the satellite.

The potential energy of satellite A is

$U_A=-\frac{GMm_A}{R+h_A}$

The potential energy of satellite B is

$U_B=-\frac{GMm_B}{R+h_B}$

So the ratio of the potential energy of satellite B to that of satellite A is

$\frac{U_B}{U_A}=\frac{m_B}{m_A}\frac{R+h_A}{R+h_B}$

The two satellites have same mass, $m_A = m_B = m$, so the equation becomes

$\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}$

where

$R=6370 km\\h_A = 7580 km\\h_B = 22000 km$

Substituting into the equation,

$\frac{U_B}{U_A}=\frac{6370 km+7580 km}{6370 km+22000km}=0.49$

(b) 0.49

The kinetic energy of a satellite in orbit is given by

$K=-\frac{1}{2}U$

where U is the potential energy.

The kinetic energy of satellite A is

$K_A=-\frac{1}{2}U_A$

The kinetic energy of satellite B is

$K_B=-\frac{1}{2}U_B$

So the ratio of the kinetic energy of satellite B to that of satellite A is

$\frac{K_B}{K_A}=\frac{U_B}{U_A}$

and we already found the value of this ratio in part (a), so this ratio is also equal to 0.49.

(c) Satellite A

The total energy of a satellite in orbit is sum of potential energy and kinetic energy:

$E=U+K=U-\frac{1}{2}U=\frac{1}{2}U$

We can write therefore the total energy of satellite A as

$E_A=\frac{1}{2}U_A$

and the total energy of satellite B as

$E_B=\frac{1}{2}U_B$

The ratio of the total energy of satellite B to that of satellite A is therefore

$\frac{E_B}{E_A}=\frac{U_B}{U_A}=0.49$

which implies that satellite A has greater total energy then satellite B.

(d) by $1.32 \cdot 10^8 J$

In order to find the difference between the total energies of the two satellite, we must calculate the exact value for each satellite. Using m = 18.2 kg and $M=5.98\cdot 10^{24}kg$ (Earth's mass), we find:

$E_A = \frac{1}{2}U_A = \frac{1}{2}\frac{GMm}{R+h_A}=\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(18.2 kg)}{6370\cdot 10^3 m+7580\cdot 10^3 m}=2.60\cdot 10^8 J$

$E_B = \frac{1}{2}U_B = \frac{1}{2}\frac{GMm}{R+h_B}=\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(18.2 kg)}{6370\cdot 10^3 m+22000\cdot 10^3 m}=1.28\cdot 10^8 J$

So the difference between the two energies is

$\Delta E=2.60\cdot 10^{8}J-1.28\cdot 10^8 J=1.32 \cdot 10^8 J$

Answer 2

The ratio of the potential energies of satellite A to that of Satellite B is $\frac{513}{1000}$.

Solution for Question A

Since the altitudes are given, it is the orbital radii that enter the equations is given as: rₐ = (7580  + 7580 )Km = 15,160Km while $r_{b}$ = (7580   + 22000) km = 29,580Km.

$\frac{Q_{b} }{Q_{a} }$ = $\frac{-GmM/Rb}{-GmM/Ra}$

= $\frac{Ra}{Rb}$ = 15,160/29,580

= 0.513 or  $\frac{513}{1000}$

(b) the ratio of kinetic energies is

$\frac{X_{b} }{X_{a} }$ = $\frac{-GmM/2Rb}{-GmM/2Ra}$

= $\frac{Ra}{Rb}$ =  15,160/29,580

= 0.513 or  $\frac{513}{1000}$

(C) Which satellite has the greater total energy?

The satellite with the greatest energy is B because It has the largest value of r.

(D) To get the variance between the energy levels of each mass we say:

Δ E = Eb - Ea; where E is energy.

= $\frac{-GmM}{2}$ $(\frac{1}{rb}) - (\frac{1}{ra})$

Therefore, Δ E = 1.1 x 10¹³J

Learn more about Gravitational Exercises at:
https://brainly.com/question/14381314