Satellite orbiting around Earth over equator

[Yoonique]

Homework Statement

A satellite is in a circular equatorial orbit around the Earth (over the equator). If the satellite flies over Singapore 9 times a day, what is the radius of the orbit?

Homework Equations

r³ = T²GM_E/4π²

The Attempt at a Solution

G = 6.67x10^-11
M_E = 5.97x10²⁴
T = (24x60x60)/9 = 9600

r = 9.76 x 10⁶m

But the answer is 2.3x10⁷m. Am I missing out anything? Or is the answer wrong?

 

[BvU]

I get the same answer as you do. Check it here

 

[gneill]

The question is a bit vague: What would happen if the orbit was retrograde rather than prograde?

 

[Yoonique]

The question is a bit vague: What would happen if the orbit was retrograde rather than prograde?

I think the period of the satellite would change, but I don't know how much would it change. If the orbit was retrograde, then the relative speed of the satellite to the Earth would be V_s + V_E rather than V_s - V_E.

 

[haruspex]

I think the period of the satellite would change, but I don't know how much would it change. If the orbit was retrograde, then the relative speed of the satellite to the Earth would be V_s + V_E rather than V_s - V_E.

gneill's point is that you have treated Singapore as a fixed point in space. While the satellite goes around the Earth at some rate w, what is Singapore doing?
That said, I can't see that it will make much difference to the answer... about 7%.

 

[BvU]

Oh boy, I completely overlooked that one !

 

[Merlin3189]

Has anyone checked the "But the answer is 2.3x107m. Am I missing out anything? Or is the answer wrong?"
That seems to me to give an orbital period of 9hrs 38min, which I can't relate to 9 passes.

 

[Yoonique]

gneill's point is that you have treated Singapore as a fixed point in space. While the satellite goes around the Earth at some rate w, what is Singapore doing?
That said, I can't see that it will make much difference to the answer... about 7%.

How do I calculate the period if I have to take into account that Singapore is also moving at w = 7.27x10^-5? I need to calculate it in terms in the frame of reference of the Earth? So angular velocity of the satellite is w_s - 7.27x10^-5 with respect to the Earth if it is in a prograde orbit?

 

[jbriggs444]

How do I calculate the period if I have to take into account that Singapore is also moving at w = 7.27x10^-5? I need to calculate it in terms in the frame of reference of the Earth? So angular velocity of the satellite is w_s - 7.27x10^-5 with respect to the Earth if it is in a prograde orbit?

The period is easy. If Singapore goes around the Earth once every 24 hours (approximately) and the satellite passes over Singapore 9 times in those 24 hours, how many times around the Earth will the satellite have gone?

It is helpful if you include units when you quote figures. The angular velocity of Singapore is approximately 7.27x10^-5 radians per second. But if you are going to quote that figure to three decimal places, you should think about using sidereal days.

 

[Yoonique]

The period is easy. If Singapore goes around the Earth once every 24 hours (approximately) and the satellite passes over Singapore 9 times in those 24 hours, how many times around the Earth will the satellite have gone?

It is helpful if you include units when you quote figures. The angular velocity of Singapore is approximately 7.27x10^-5 radians per second. But if you are going to quote that figure to three decimal places, you should think about using sidereal days.

I calculated if they both orbit in the same direction, ω of satellite is 10x faster than ω of Singapore. So the period of the satellite is 10x less than the period of Singapore.
However is they both orbit in different direction, ω of satellite is 8x faster than ω of Singapore. So the period of the satellite is 8x less than the period of Singapore. Both answers are less than the given answer by a magnitude of 10. I assume the given answer is wrong?

 

[jbriggs444]

Yes, the given answer is wrong.

Sanity check on your work:

Gravitational force scales as 1/r². Orbital acceleration scales as ω²r. Set those two equal (ignoring the constants of proportionality) and you have 1/r² = ω²r. Divide by r and you get 1/r³ = ω²

If we want to multiply ω by 10, that means that ω² is up by a factor of 100 and r must be down by a factor of the cube root of 100. That's a factor somewhere between 4 and 5. [4 cubed is 64 and 5 cubed is 125]

Geosynchronous orbit is about 40,000 km up from the surface of the earth. Or 46,000 km up from the Earth's center. Divide that by a factor of 4 or 5 and you have 10,000 km up from the center. That matches your answer very nicely.

Sanity check on the given answer:

The given answer is 23 million meters = 23,000 km. That's almost exactly half of the geosynchronous radius of 46,000 km. One might speculate that the question was prepared by someone who figured a retrograde orbit (factor of 8 rather than the correct factor of 10), correctly computed a relationship to the cube of the radius (factor of 8 now down to a factor of 2) and then forgot to take the square root to account for the squared dependence on angular velocity.

 

[Yoonique]

Yes, the given answer is wrong.

Sanity check on your work:

Gravitational force scales as 1/r². Orbital acceleration scales as ω²r. Set those two equal (ignoring the constants of proportionality) and you have 1/r² = ω²r. Divide by r and you get 1/r³ = ω²

If we want to multiply ω by 10, that means that ω² is up by a factor of 100 and r must be down by a factor of the cube root of 100. That's a factor somewhere between 4 and 5. [4 cubed is 64 and 5 cubed is 125]

Geosynchronous orbit is about 40,000 km up from the surface of the earth. Or 46,000 km up from the Earth's center. Divide that by a factor of 4 or 5 and you have 10,000 km up from the center. That matches your answer very nicely.

Sanity check on the given answer:

The given answer is 23 million meters = 23,000 km. That's almost exactly half of the geosynchronous radius of 46,000 km. One might speculate that the question was prepared by someone who figured a retrograde orbit (factor of 8 rather than the correct factor of 10), correctly computed a relationship to the cube of the radius (factor of 8 now down to a factor of 2) and then forgot to take the square root to account for the squared dependence on angular velocity.

Thanks this is a nice strategy to check for answers when dealing with large numbers, will take note of this.