Space physics, is this correct?

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A satellite in orbit above Earth's equator is traveling at an orbital speed of 7.45 km/s, leading to calculations for its altitude and period. The initial calculations yielded a radius of approximately 7.1 million meters, suggesting an altitude of around 700 km after subtracting Earth's radius. However, further discussions revealed that a more accurate altitude is closer to 810 km due to more precise values for Earth's mass. The period was calculated using the formula T = 2πr/V, resulting in approximately 5988 seconds. Overall, the calculations were deemed correct, with emphasis on the importance of precision in values used.
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Homework Statement


A satellite in orbit above Earths equator is traveling at an orbital speed of 7.45 km/s.

a) Determine the altitude of the satellite
b) Determine the satellites period


Homework Equations


v = sqrt(Gm/r)

The Attempt at a Solution


7.45 * 1000 = 7450 m/s

I plugged in the values to that equation and got a radius of 7.1 * 10^6 m.

Now, I'm not sure how to find the altitude, but shouldn't it be the value I found (7.1*10^6m) minus Earth's radius? This gives me the value of 700 km

For b, I'm not sure how to find the period, I tried T = 2pir / V where r was the radius I found and got 5988 seconds. Is this correct?
 
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x86 said:

Homework Statement


A satellite in orbit above Earths equator is traveling at an orbital speed of 7.45 km/s.

a) Determine the altitude of the satellite
b) Determine the satellites period


Homework Equations


v = sqrt(Gm/r)

The Attempt at a Solution


7.45 * 1000 = 7450 m/s

I plugged in the values to that equation and got a radius of 7.1 * 10^6 m.

Now, I'm not sure how to find the altitude, but shouldn't it be the value I found (7.1*10^6m) minus Earth's radius? This gives me the value of 700 km

For b, I'm not sure how to find the period, I tried T = 2pir / V where r was the radius I found and got 5988 seconds. Is this correct?

All of those are approximately correct. You are doing the right things. I get a distance closer to 800km. If you want a closer check you'll need to show all the numbers you are using.
 
Dick said:
All of those are approximately correct. You are doing the right things. I get a distance closer to 800km. If you want a closer check you'll need to show all the numbers you are using.

okay, thank you for your help. I am doing things this way:

a)
7450 = sqrt((6.67*10^-11 * 5.9*10^24/(r)

r = 700 km
 
EDIT: Didn't see this had already been solved; ignore this post.
 
x86 said:
okay, thank you for your help. I am doing things this way:

a)
7450 = sqrt((6.67*10^-11 * 5.9*10^24/(r)

r = 700 km

Nah, that doesn't give you r=700 km. That gives you the radius of the orbit. You have to subtract the radius of Earth to get anything close to 700 km. What did you get for that r, and then what did you use for the radius of earth? As I already said, I think you are doing everything correctly except for maybe rounding off and exact numbers. I really wouldn't worry about it.
 
Last edited:
Dick said:
Nah, that doesn't give you r=700 km. That gives you the radius of the orbit. You have to subtract the radius of Earth to get anything close to 700 km. What did you get for that r, and then what did you use for the radius of earth?

Ah yes, I was rounding the mass of the Earth too much. I get about 810 km now after using a different value of the mass. Thank you
 
x86 said:
Ah yes, I was rounding the mass of the Earth too much. I get about 810 km now after using a different value of the mass. Thank you

Right. You were doing everything correctly. You should pay attention to round-off etc. Very welcome.
 
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