# Satellites Orbits and Energy Problem

## Homework Statement

Two small spaceships, each with mass m = 2000 kg, are in the circular Earth orbit of the figure, at an altitude h of 440 km. Igor, the commander of one of the ships, arrives at any fixed point in the orbit 81.0 s ahead of Picard, the commander of the other ship. What are the (a) period T0 and (b) speed v0 of the ships? At point P in the figure, Picard fires an instantaneous burst in the forward direction, reducing his ship's speed by 2.00%. After this burst, he follows the elliptical orbit shown dashed in the figure. What are the (c) kinetic energy and (d) potential energy of his ship immediately after the burst? In Picard’s new elliptical orbit, what are (e) the total energy E, (f) the semimajor axis a, and (g) the orbital period T? (h) How much earlier than Igor will Picard return to P?

## Homework Equations

kepler's 3rd law: T^2=4*(π^2)*(r^3)/(GM)
Newton's law: F=GMm/R^2
Centripetal Force: F=mv^2/R
1st Escape Speed= v=sqrt(GM/R)
KE=0.5mv^2=GMm/2R ------- by substitute v into the equation
GPE=-GMm/R
E=KE+GPE=-KE

where M=5.972*10^24 kg and m=2000kg

## The Attempt at a Solution

For (a), I used keeper's law T^2=4*(π^2)*(r^3)/(GM) and took the sqrt of RHS, got T=91.88s
I got this one wrong
b) i used mv^2/R=GMm/R^2
and got the expression of v=sqrt(GM/R) ----- or just take the escape speed
v=3.01*10^4 again WRONG
c)KE=0.5m(v')^2=8.69*10^11 ----- where v'=0.98v
again this is WRONG solution
d)GPE=-2KE or GPE=-GMm/R
sub in values got GPE=-1.74*10^12
e)E=KE+GPE=-KE=-8.69*10^11
now the rest depends on the correct solutions of previous ones, I wouldn't bother typing in the rest of my attempt. Please enlighten me with your genius minds

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SteamKing
Staff Emeritus
Homework Helper

## Homework Statement

Two small spaceships, each with mass m = 2000 kg, are in the circular Earth orbit of the figure, at an altitude h of 440 km. Igor, the commander of one of the ships, arrives at any fixed point in the orbit 81.0 s ahead of Picard, the commander of the other ship. What are the (a) period T0 and (b) speed v0 of the ships? At point P in the figure, Picard fires an instantaneous burst in the forward direction, reducing his ship's speed by 2.00%. After this burst, he follows the elliptical orbit shown dashed in the figure. What are the (c) kinetic energy and (d) potential energy of his ship immediately after the burst? In Picard’s new elliptical orbit, what are (e) the total energy E, (f) the semimajor axis a, and (g) the orbital period T? (h) How much earlier than Igor will Picard return to P?

## Homework Equations

kepler's 3rd law: T^2=4*(π^2)*(r^3)/(GM)
Newton's law: F=GMm/R^2
Centripetal Force: F=mv^2/R
1st Escape Speed= v=sqrt(GM/R)
This is not escape speed. It is orbital speed.
KE=0.5mv^2=GMm/2R ------- by substitute v into the equation
GPE=-GMm/R
E=KE+GPE=-KE

where M=5.972*10^24 kg and m=2000kg
It's not clear from your calculations what number you are using for R.

## The Attempt at a Solution

For (a), I used keeper's law T^2=4*(π^2)*(r^3)/(GM) and took the sqrt of RHS, got T=91.88s
I got this one wrong
I should say so. The period for objects in low earth orbit is typically on the order of 90 minutes, rather than 90 seconds.
b) i used mv^2/R=GMm/R^2
and got the expression of v=sqrt(GM/R) ----- or just take the escape speed
Again, not the escape speed - it's the orbital speed. There's a difference. What value are you using for R?
v=3.01*10^4 again WRONG
c)KE=0.5m(v')^2=8.69*10^11 ----- where v'=0.98v
again this is WRONG solution
d)GPE=-2KE or GPE=-GMm/R
sub in values got GPE=-1.74*10^12
e)E=KE+GPE=-KE=-8.69*10^11
now the rest depends on the correct solutions of previous ones, I wouldn't bother typing in the rest of my attempt. Please enlighten me with your genius minds

• i_hate_math
This is not escape speed. It is orbital speed.

It's not clear from your calculations what number you are using for R.

I should say so. The period for objects in low earth orbit is typically on the order of 90 minutes, rather than 90 seconds.

Again, not the escape speed - it's the orbital speed. There's a difference. What value are you using for R?
I used R=440km=4400000m for R but Ive now realised that R should be the radius of the earth plus altitude that is R=6371+440=6811km=6811000m
Was my expression for orbital speed correct? apart from my false understanding of escape speed?

This is not escape speed. It is orbital speed.

It's not clear from your calculations what number you are using for R.

I should say so. The period for objects in low earth orbit is typically on the order of 90 minutes, rather than 90 seconds.

Again, not the escape speed - it's the orbital speed. There's a difference. What value are you using for R?
Also I have reworked my solution with the new R, I got 5.60*10^3 sec for T, which is close to the 90min you mentioned

SteamKing
Staff Emeritus
Homework Helper
I used R=440km=4400000m for R but Ive now realised that R should be the radius of the earth plus altitude that is R=6371+440=6811km=6811000m
Was my expression for orbital speed correct? apart from my false understanding of escape speed?
Yes, your formula is correct for a circular orbit.

• i_hate_math
The equation GPE=-2KE should be replaced with GPE=-GMm/R since KE in this case is not conserved

gneill
Mentor
@i_hate_math: I see that this thread has been marked solved. Has it really been solved to your satisfaction?

@i_hate_math: I see that this thread has been marked solved. Has it really been solved to your satisfaction?
Yes @SteamKing has been very helpful