Satellites Orbits and Energy Problem

In summary, the two ships have a period of 91.88 seconds and a speed of 3.01*10^4 meters per second when they reach point P in the Earth's orbit. The ship that arrives first returns to point P earlier than the ship that arrives later.
  • #1
i_hate_math
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2

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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  • #2
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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  • #3
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?
 
  • #4
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
 
  • #5
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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  • #6
The equation GPE=-2KE should be replaced with GPE=-GMm/R since KE in this case is not conserved
 
  • #7
@i_hate_math: I see that this thread has been marked solved. Has it really been solved to your satisfaction?
 
  • #8
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
 

Related to Satellites Orbits and Energy Problem

What is a satellite?

A satellite is an object that orbits around a larger object, such as a planet or a star. It can be natural, like the moon, or man-made, like a communication satellite. Satellites are often used for communication, weather forecasting, navigation, and scientific research.

How do satellites maintain their orbits?

Satellites maintain their orbits through the balance of two forces: gravity and centrifugal force. Gravity pulls the satellite towards the center of the object it is orbiting, while centrifugal force pushes it away. These forces create a circular or elliptical path around the object.

What is the energy problem faced by satellites in orbit?

The energy problem faced by satellites in orbit is the gradual decrease in their orbital energy over time. This is due to several factors, such as atmospheric drag, solar radiation pressure, and gravitational interactions with other objects in space. This decrease in energy can cause satellites to fall out of orbit or lose their functionality.

How is the energy problem solved for satellites in orbit?

To solve the energy problem, satellites often have thrusters or small rockets that can be used to adjust their orbits and counteract the effects of atmospheric drag and other forces. Additionally, satellites can be placed in higher orbits where there is less atmospheric drag, or they can be equipped with solar panels to generate energy and counteract the effects of solar radiation pressure.

What are the different types of satellite orbits?

There are several different types of satellite orbits, including low Earth orbit, medium Earth orbit, and geostationary orbit. Low Earth orbit is the closest to Earth and is used for imaging and communication satellites. Medium Earth orbit is slightly higher and is used for navigation and communication satellites. Geostationary orbit is the highest and is used for weather and communication satellites that need to maintain a fixed position relative to Earth.

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