Satellites Orbits and Energy Problem

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Homework Help Overview

The problem involves two small spaceships in circular orbit around the Earth, with a focus on calculating their orbital period, speed, and energy changes after a speed reduction. The context includes gravitational physics and orbital mechanics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of Kepler's laws and Newton's laws to find the period and speed of the spaceships. There are attempts to calculate kinetic and potential energy after a speed change. Some participants question the use of escape speed versus orbital speed.

Discussion Status

There is ongoing exploration of the correct values for radius and the implications for calculations. Some participants have provided clarifications on the differences between escape speed and orbital speed, and there is acknowledgment of the need for accurate radius values in calculations. Guidance has been offered regarding the correct formulas for potential energy.

Contextual Notes

Participants note discrepancies in the values used for the radius of orbit, with discussions about the correct total radius including Earth's radius plus altitude. There is also mention of typical periods for low Earth orbit, which suggests a need for reevaluation of initial calculations.

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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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i_hate_math said:

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
 
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SteamKing said:
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 I've now realized 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?
 
SteamKing said:
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
 
i_hate_math said:
I used R=440km=4400000m for R but I've now realized 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.
 
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The equation GPE=-2KE should be replaced with GPE=-GMm/R since KE in this case is not conserved
 
@i_hate_math: I see that this thread has been marked solved. Has it really been solved to your satisfaction?
 
gneill said:
@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
 

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