Radius of Orbits: GPS Satellite & Earth

[Ryo124]

Homework Statement

A GPS satellite is put in a high circular orbit around the earth. The period of revolution is 8 hours. Calculate the radius r of the orbit, compared to the radius R of the earth.

Homework Equations

Rearth= 6.4 x 10^6 m
mearth= 6 x 10^24 kg

The Attempt at a Solution

I have no idea how to attempt this problem. Can someone please help?

 

[Dick]

Write down the acceleration of an object moving in circular orbit of radius r with period T. Equate that to the acceleration caused by gravity at radius r.

 

[Ryo124]

It is not clear to me what you mean. Don't you mean velocity? How does the mass of the Earth come into play? So, step-by-step, what do I do? Whats first? Please Help.

 

[Dick]

No, I mean acceleration. An object moving in a circular path experiences a centripetal acceleration. There's a formula for it in terms of v and r. Now equate that to the gravitational acceleration. There's a formula for that as well. Find those formulas and set them equal.

 

[Ryo124]

Ohhh... I see now that it is ac. mv^2/r right? I'll try to work it out and respond if i get stuck, thanks.

 

[nrqed]

Ohhh... I see now that it is ac. mv^2/r right? I'll try to work it out and respond if i get stuck, thanks.

Right (m is the mass of the satellite). Now set this equal to the force of gravity of the Earth on the satellite (using the universal law of gravitation). You will also need to use that for circular motion, v = \frac{2 \pi r }{ T}

 

[Ryo124]

v = \frac{2 \pi r }{ T}

Do I find the velocity for Earth's orbit AND the satellite with this equation? Or just the satellite? Isn't the mass of the satellite negligible, so set it to 1?

 

[Dick]

I think finding the radius of the satellite is the question. You can set the mass of the satellite equal to one if you wish, it will cancel from both sides of the equation you have yet to write down. And sure, you'll want to assume it's negligible compared with the Earth so you can assume the Earth is stationary while the satellite orbits. WRITE DOWN THE EQUATION, ok?

 

[Ryo124]

mv^2/r = Gm1m2/R^2

Is this the equation? And all values are known except r, correct? And you find the v from v = \frac{2 \pi r }{ T} correct?

So this would mean that the centripetal force from the satellite equals the gravitational force between the Earth and the satellite??

 

[Dick]

Finally. Yes. One of the m1 and m2 on the right is equal to m on the left. Cancel it. The remaining m is mass of the earth, right?

 

[Ryo124]

How are you supposed to find the v of the satellite using v = \frac{2 \pi r }{ T} if you don't know what r is?

 

[D H]

From post #9, you have v^2/r=GM_e/r^2. You don't know v but you do have an expression for it: v=2\pi r/T. So substitute this expression in the equation from post #9 and solve for r.

 

[Ryo124]

I keep getting 8.08 x 10^14 m for r, but this is incorrect. Is my math just wrong? You set [2(pi)r/(T)]^2 = GME/RE^2 right?

 

[Dick]

No. i) the quantity on the left should be v^2/r. I think you are missing the r. And the R_E on the right should be r of the satellite, right?

 

[Ryo124]

Sorry. When I did the calculation I had the r on the left side and the right side, but I still keep getting the answer incorrect.

The equation is v^2/r=GM_e/r^2 with v = \frac{2 \pi r }{ T}

I know this and keep getting the wrong answer of 8.08 x 10^14. What are you getting??

 

[Dick]

I get something round about 10^7m. Can you post your final equation for r and maybe some intermediate results?

 

[Dick]

You should be getting that r^3 is proportional to T^2. That's one of Kepler's laws.

 

[Ryo124]

I am getting [2(pi)r]^2 x [1/T]^2 = GME/r^2

From here I can't simplify and I am not getting 10^7. I'm getting around 10^14 instead.

When I tried to simplify, I got [2(pi)r]/(8.29x10^8) = [4.002x10^36]/(r^2). Then I crossed multiplied and got the wrong answer.

I am not simplifying correctly, can someone help me with this, I feel dumb.

 

[Dick]

The 1/r in v^2/r has disappeared again. You should be able to simplify it to r^3=(something)*T^2. What's the (something)?

 

[Ryo124]

Is it GME/ 2(pi) ?

 

[Dick]

Would you believe GM_E/(2pi)^2??

 

[Ryo124]

I plug that in and get 5.5 x 10^14 m, which is wrong.

 

[D H]

Show us exactly what you did. Believe it or not, Dick's expression is quite correct.

 

[Ryo124]

OK. Centripetal force from the satellite equals the gravitational force between the Earth and the satellite:

mv^2/r = Gm1m2/r^2

The mass of the satellite is negligible:

v^2/r = Gme/r^2

The velocity in terms of radius and period is 2(pi)r/T so:

[2(pi)r/T]^2/r = Gme/r^2

An r cancels on the left side and you have:

[2(pi)^2(r)]/T^2 = Gme/r^2

Cross multiply and get:

[2(pi)^2]r^3 = GmeT^2

I plug in the numbers and get 5.5 x 10^14 m, which is wrong.

 

[Dick]

Tell us what numbers you are plugging in where. Otherwise, are we supposed to figure out why you are getting a wrong number?

 

[Ryo124]

For G i put 6.67 x 10^11 G = 6.67 x 10^11
For Me i put 6 x 10^24 kg Me = 6.0 x 10^24 kg
For T i put 28800 sec T = 28800 sec

What am I doing wrong? Are my steps correct? What are you getting for your answer?

 

[D H]

One error: you didn't square the entire expression, 2\pi r/T. However, correcting this will not fix the 10¹⁴ problem. As Dick noted, please show exactly what numbers you used.

 

[nrqed]

For G i put 6.67 x 10^11 G = 6.67 x 10^11
For Me i put 6 x 10^24 kg Me = 6.0 x 10^24 kg
For T i put 28800 sec T = 28800 sec

G is 6.67 times 10 to the minus 11

 

[D H]

G is 6.67 times 10 to the minus 11

Even more correctly, 6.67x10^-11 m³/kg/s². It is a good idea to always explicitly represent units.

 

[nrqed]

Even more correctly, 6.67x10^-11 m³/kg/s². It is a good idea to always explicitly represent units.

You are right. I just wanted to emphasize the problem with the exponent.

 

[D H]

You're right too. That is a very big problem.

 

[Ryo124]

Alright, I changed it and got 2.56 x 10^7 m for the radius of the orbit. . . Can someone work the problem out and see if this is the correct answer?

Also, no one told me if my final equation was correct... [2(pi)^2]r^3 = GmeT^2
Is this equation correct?

 

[nrqed]

Alright, I changed it and got 2.56 x 10^7 m for the radius of the orbit. . . Can someone work the problem out and see if this is the correct answer?

Also, no one told me if my final equation was correct... [2(pi)^2]r^3 = GmeT^2
Is this equation correct?

I get 2.03 \times 10^7 m You forgot to square the 2. It's (2 pi)^2

 

[Ryo124]

OK. I did it and got 2.04 x10^7 m. I did you just round up in your calculation?

 

[nrqed]

OK. I did it and got 2.04 x10^7 m. I did you just round up in your calculation?

I keep getting 2.03 x 10^7 m. I did not round off.
(I mean I rounded off my final answer but not in the intermediate steps)

 

[Ryo124]

OK. Weird, I did it again and got 2.03 x 10^7. Cool.

Thanks for your help and patience everyone, I guess I made this problem harder than it needed to be.

 

[D H]

Since you only used two decimal places of accuracy for G and M_e, it would be better to say 2.0x10⁷ meters in this case.

As an aside, it is often better to use the planet gravitational constant (the product of universal gravitational constant and the planet mass rolled into one constant) rather than the product of universal gravitational constant and the planet mass as separate values. What's the difference? Accuracy. In the case of the Earth, we know the product G M_e to 9 places of accuracy: 398,600.4418±0.0008 km³/s². In comparison, we only know G to 4 decimal places.