Potential Energy and raising a satellite from Earth into a Circular Orbit

[AN630078]

Homework Statement

Hello, I have quite a lengthy, but assuredly related question where each section leads onto the next. I have been trying to better understand potential energy and the mechanisms of circular orbits when I came across the problems below. Would anyone be able to scrutinise my calculations and evaluate my workings, to perhaps suggest a preferable approach. For instance in hindsight would it have been better if I had applied Newton's Second Law, ΣF = ma,
which becomes GmM/r2 = mv2/r. Then continued to derive formula in terms of potential and kinetic energy? Thank you to anyone who replies, apologies for the length, although they are related questions, ie to find the kinetic energy one must first establish the value of the orbital velocity.

The radius of the Earth is 6.4 x 10^6 m and its mass is 6.0 x 10^24 kg.

Question 1:
a.What is the gravitational potential at the surface of the Earth?
b.What is the gravitational potential 200 km above the Earth’s surface?
c. How much energy would be needed to move a satellite of mass 3000 kg from the Earth’s surface to a height of 200 km?
d. At what speed would a satellite 200 km above the Earth’s surface be moving?
e. If its mass were 3000 kg how much kinetic energy would it have?
f. What total energy is needed to set a satellite of mass 3000 kg in orbit 200 km above the Earth?

Relevant Equations

V grav= -GM/r
E=U+K
v=√GM/r

a. V=-GM/r
V=-6.67*10^-11*6.0 x 10^24/6.4 x 10^6
V grav = -62531250 ~ -62.5M Jkg^-1

b. To find the gravitational potential 200 km above the surface of the Earth;
r=6.4 x 10^6 +2*10^5 m=6.6*10^6
V grav=-6.67*10^-11*6.0 x 10^24/6.6*10^6
V grav= -60636363 ~ -60.6 M Jkg^-1

Can I check that it is correct that these values are negative?

c. To find the energy required to move a satellite to a height of 200km above the surface of the Earth find the difference in the gravitational potential;
ΔU=Uorbit−UEarth

U Earth = -GMm/r
U Earth = - 6.67*10^-11*6.0 x 10^24*3000/ 6.4 x 10^6
U Earth = -1.8759375 * 10^11 J~ -1.87* 10^11 J

U orbit = -GMm/r + 2.0*10^5 m
U orbit = - 6.67*10^-11*6.0 x 10^24*3000/ 6.6 x 10^6
U orbit = -1.819090 *10^11 J ~ -1.82 *10^11 J

ΔU=( -1.82 *10^11)-( -1.8759375 * 10^11)
ΔU=5684659091 ~ 5700000000 J = 5700 MJ

Are units of J or MJ suitable her or should it be Jkg^-1?

d. To find the orbital velocity use the formula;
v orbit = √GM/r
v orbit = √ 6.67*10^-11*6*10^24/6.6*10^6
v orbit =7786.935446 ~ 7790 ms^-1 to 3.s.f

e.The kinetic energy of the satellite on the surface of the Earth is 0 J.
The kinetic energy of the satellite 200 km above the Earth's surface is;
K orbit = 1/2mv^2
K orbit = 1/2 * 3000*7790^2
K orbit = 9.1 * 10 ^10 J to 2.s.f
KE total = K Earth + K orbit
KE total = 0 J + 9.1 * 10 ^10 J=9.1 * 10 ^10 J

This can be verified by subtracting the change in potential energy from the total energy.

f. E total=K1+U1=K2+U2
The total energy required is the difference in the satellite's energy in orbit and that at Earth’s surface.
E orbit = K orbit + U orbit
E orbit= 9.1 * 10 ^10 + (-1.819 *10^11)= -9.1 * 10^10 J

E surface = K Earth + U Earth
E surface = 0 + (-1.87* 10^11)=-1.87* 10^11 J

Δ𝐸= E orbit - E surface
Δ𝐸=-9.1 * 10^10-(-1.87* 10^11)
Δ𝐸=9.66392 * 10^10 ~ 9.66 *10^10 J

 

[haruspex]

Re b), calculating the additional GPE of GPE_200km-GPE₀ by taking the difference of those two large numbers tends to produce a significant error in the answer. To avoid keeping lots of digits, it is better to do a bit of algebra first:
$\frac{GMm}{r_{200}}-\frac{GMm}{r_0}=-GMm(\frac{1}{r_{0}}-\frac{1}{r_{200}})=-\frac{GMm}{r_0r_{200}}200$

 

[AN630078]

Re b), calculating the additional GPE of GPE_200km-GPE₀ by taking the difference of those two large numbers tends to produce a significant error in the answer. To avoid keeping lots of digits, it is better to do a bit of algebra first:
$\frac{GMm}{r_{200}}-\frac{GMm}{r_0}=-GMm(\frac{1}{r_{0}}-\frac{1}{r_{200}})=-\frac{GMm}{r_0r_{200}}200$

Thank you for your reply. Yes I have seen this formula before but I was unsure how to derive it. Thank you for the suggestion Are there any other improvements I could make?

 

[haruspex]

The kinetic energy of the satellite on the surface of the Earth is 0 J.

Only at the poles.

 

[AN630078]

Only at the poles.

Thank you again so would the rest of my workings be correct also? I will use the formula you suggested in b also, thank you again for providing that

 

[haruspex]

Thank you again so would the rest of my workings be correct also? I will use the formula you suggested in b also, thank you again for providing that

Yes, it's fine.