# Universal gravitation 1-determine height of satellite

1. Jun 14, 2012

### dani123

1. The problem statement, all variables and given/known data

Two satellites are orbiting around the Earth. One satellite has a period of 1.5 h and is 250 km above the Earth's surface. The second satellite has a period of 7.5 h. Using Kepler's Laws and the fact that the Earth's radius is 6.38x106 m determine the height of the second satellite above the Earth's surface.

2. Relevant equations

Kepler's 3rd law: (Ta/Tb)2=(Ra/Rb)3

motion of planets must conform to circular motion equation: Fc=4∏2mR/T2

From Kepler's 3rd law: R3/T2=K or T2=R3/K

Gravitational force of attraction between the sun and its orbiting planets: F=(4∏2Ks)*m/R2=Gmsm/R2

Gravitational force of attraction between the Earth and its orbiting satelittes: F=(4∏2Ke)m/R2=Gmem/R2

Newton's Universal Law of Gravitation: F=Gm1m2/d2

value of universal gravitation constant is: G=6.67x10-11N*m2/kg2

weight of object on or near Earth: weight=Fg=mog, where g=9.8 N/kg
Fg=Gmome/Re2

g=Gme/(Re)2

determine the mass of the Earth: me=g(Re)2/G

speed of satellite as it orbits the Earth: v=√GMe/R, where R=Re+h

period of the Earth-orbiting satellite: T=2∏√R3/GMe

Field strength in units N/kg: g=F/m

Determine mass of planet when given orbital period and mean orbital radius: Mp=4∏2Rp3/GTp2

3. The attempt at a solution

So for satellite #1 we have,
T1=1.5 hr
h1=250 km above earth's surface

For second satellite we have,
T2=7.5hr
h2=?

We know that the earth's radius is RE=6.38x106 m

so R1=250000m+ (6.38x106m)= 6630000m

R2=RE+h2

I used (Ta/Tb)2=(Ra/Rb)3
to solve for R2=8.098x1010m

Then I did, R2-RE=h2
h2=80 970 595 km

Could someone please just verify that what I did here is correct? and if I made any mistakes if someone could please point them out to me that would be greatly appreciated! Thanks so much in advance !

2. Jun 14, 2012

### SammyS

Staff Emeritus
I crossed out a bunch of stuff you don't need for this problem.

You have definitely done something wrong! The distance from Sun to Earth is only about 1.5×108km .

The ratio of T2/T1 = 5.

The ratio of R2/R1 should be less than 5.

Did you forget to take the cube root in getting your answer?

3. Jun 14, 2012

### dani123

You are absolutely right, i didn't put it in my calculator correctly... so this time i got R2=8998.29 km or 8998294.4m

but from here, how do i go about getting the value for h2 which is ultimately what the question is asking for?

4. Jun 14, 2012

### dani123

I proceeded to do the exact thing as I did in my attempt above and got h2=2618.29 km

Does this seem reasonable?

Thanks again so much for your help!

5. Jun 14, 2012

### SammyS

Staff Emeritus
I don't think that's right.

Show your steps in detail.

6. Jun 15, 2012

### dani123

I did:

R1=(250000m)+(6.38x106m)= 6 630 000m

Then I went on to find R2=Re+h

So with the equation (T1/T2)2=(R1/R2)3

(R2)3= (R1)3/(T1/T2)2

R2=(6 630 000m)3/(1.5/7.5)2= 8998294.399 m

And then I manipulated the R=Re+h equation and solved for h=2618294.399 m

Last edited: Jun 15, 2012
7. Jun 15, 2012

### HallsofIvy

Staff Emeritus
The problem asks for the height of the satellite above the earth's surface. Aren't you calculating the distance from the earth's center?

8. Jun 15, 2012

### dani123

sorry, yes your right, I just fixed it... does that seem right?

9. Jun 23, 2012

### dani123

does h= 2618294.399m seem like the correct answer?