Circular Orbits of Satelites and Planets

[.NoStyle]

Homework Statement

A spy satellite is in circular orbit around Earth. It makes one revolution in 6.00 h.

A. How high above Earth's surface is the satellite?

B. What is the satellite's acceleration?

Homework Equations

I'm not sure but I believe we could possibly derive something from

SUMofF = (Gm1m2)/r^2

and with NSL, v=sqrtof (GMe/r)

where Me is the mass of the Earth 5.974e24 kg. G is the universal gravitational constant 6.67e-11N * m^2/kg^2

The Attempt at a Solution

and r is the radius of the Earth plus the radius of the height. So we can find the height that the satellite is by deducting the radius of the Earth (6.371e6 m) from r to find the height.

I don't know what equation to use and I don't know how I would find the acceleration.


Thanks for your help!

 

[rock.freak667]

The force of attraction between the Earth and satellite,provides the force needed to keep the satellite in a circular orbit.

 

[.NoStyle]

would it be the Universal Gravitational Constant? Sorry, I'm not really clear with what you're saying. Thanks

 

[dynamicsolo]

If you set the gravitational force between Earth and satellite, given by

SUMofF = (Gm1m2)/r^2

equal to the centripetal force acting on the satellite in its circular orbit (i.e., gravity is what provides the centripetal force), and do some algebra, you'll come up with

v=sqrt (GMe/r)

This is called the "circular velocity" of the spacecraft in its orbit. If you now consider that the period, T, of the satellite is the time it takes for the spacecraft to complete the circular orbit of radius R at the speed v given above, you can do some more algebra to find a relation between T and R (you will, in fact, have found a version of Kepler's Third Law!).

The gravitational force equation will give you the satellite's acceleration in its orbit, if you keep in mind Newton's Second Law (F = ...?). [Since what they are asking for is also the centripetal acceleration of the satellite, you can use that equation and the orbital speed you found. You should -- you'd better! -- get the same answer either way...]

 

[.NoStyle]

Hi again dynamic solo, thanks for all the help lately. Before I go further, r in the "circular velocity" equation is the radius of earth? thanks

 

[rock.freak667]

Hi again dynamic solo, thanks for all the help lately. Before I go further, r in the "circular velocity" equation is the radius of earth? thanks

r is the distance between the Earth and the satellite.

 

[.NoStyle]

r is the distance between the Earth and the satellite.

Thank you. So for V, that would be 2pi/6h?

 

[dynamicsolo]

What is the distance around the circular orbit? That is the distance you are covering in those six hours.

 

[.NoStyle]

What is the distance around the circular orbit? That is the distance you are covering in those six hours.

The distance is 2pi, isn't that right? so 2pi/6h?

 

[dynamicsolo]

The circumference of a circle is given by what formula?

(The number pi has no units, so 2·pi can't be the length of anything...)

 

[.NoStyle]

hi dynamic,

the circumference is 2pi*radius

but we don't know the radius. So I have2pi*radius/6hours = sqrtof(GMe/r)

solving for r I get:r ^1.5 = sqrt of (GMe) * 6 all over 2pioh god, what am I doing?

 

[dynamicsolo]

OK, the circular speed is given by the distance around the circular orbit (circumference) divided by the time required to do so (period), so you get

2pi*r/6hours = sqrtof(GMe/r) .

But watch your algebra! You have r on one side and a square root of r in the denominator of the other side. Square both sides (also convert the period into seconds, since we'll need to work in SI metric) to get

[ (2·pi) · r ]^2 / (6 hours converted to seconds)^2
= G · Me / r .

Now solve this for r , which is the radius of the orbit. (Actually, what you had is close, but r^1.5 will be more of a nuisance to work with...)

 

[.NoStyle]

hey dynamic, once again, thanks a lot for the help.

I got it right I think!

1.039E7 meters!

hehehe

 

[dynamicsolo]

...or about 10,400 km. altitude.

Now how about part (b)?

 

[.NoStyle]

hmmm i needed help with that,

I got 8.7E-1 meters/secondthat's wrong huh?

thanks

 

[dynamicsolo]

Well, acceleration has units of meters/(second^2), so something didn't come out right...

You should probably show how you calculated this. In the meantime, here are two approaches you can try (there's a third, but I'll describe it afterwards):

You know that the force on the satellite is F = ma , so you could compute either

1) the gravitational force between the satellite and the Earth, set that equal to ma , and solve for a

or

2) find the centripetal force on the satellite (you have the radius of the orbit and the speed of the satellite along it) and set that equal to ma, and solve for a .

Notice that you don't need to know the satellite's mass m, since it will divide out (which tells us that any satellite which is small in mass compared to Earth would behave the same way).

 

[.NoStyle]

Hi Dynamicsolo,

yes it is 8.7E-1 m/s^2

this is what I did.

I used the formula asubr = omegasquared*radius

i found the requency and plugged that into the equation

arrived with that answer. The problem is, I don't know if I did it 100% correctly.

I'll try your method and report back. Thank you greatly

 

[.NoStyle]

I'm a bit confused with steps 1 and 2 since there are many names for the same thing. First off, what do you mean "gravitational force" is that the same thing as Fg = Gm1m2/r^2??and with what you're saying, we don't need to know the mass of the satellite therefore:

Fg= GMe/r^2 ?

and then plugging in the universal constant and multiplying it with the mass of Earth and dividing everything by the radius of the orbit squared will give us the gravitational force? Thanksedit -

and since you said gravitational force = m * a

what is the value of m? which mass is it? thanks

 

[dynamicsolo]

yes it is 8.7E-1 m/s^2

this is what I did.

I used the formula asubr = omegasquared*radius

i found the requency and plugged that into the equation

arrived with that answer.

Ah, that form for the centripetal acceleration will work too. It also looks like you found the angular frequency of the satellite on its orbit correctly. But, did you use the radius of the orbit, or the altitude...?

 

[dynamicsolo]

I'm a bit confused with steps 1 and 2 since there are many names for the same thing. First off, what do you mean "gravitational force" is that the same thing as Fg = Gm1m2/r^2??

That is the equation for the gravitational force between two masses.

and with what you're saying, we don't need to know the mass of the satellite therefore:

Fg= GMe/r^2 ?

You left out one of the masses this time. The gravitational force between the Earth (mass: Me) and the satellite (call its mass m) will be

F_g = G · Me · m / (r^2) ,

which will equal the centripetal force on the satellite (since the gravitational force is what is providing that force)

F_g = F_c = m · a_c ,

the centripetal acceleration being what you are solving for. The mass m appears on both sides of the equation, so it divides out (meaning that, for satellites with masses that are small compared to Earth, the mass doesn't matter -- we must modify this approach to deal with, say, the Moon).

 

[BobG]

I'm a bit confused with steps 1 and 2 since there are many names for the same thing. First off, what do you mean "gravitational force" is that the same thing as Fg = Gm1m2/r^2??


and with what you're saying, we don't need to know the mass of the satellite therefore:

Fg= GMe/r^2 ?

and then plugging in the universal constant and multiplying it with the mass of Earth and dividing everything by the radius of the orbit squared will give us the gravitational force? Thanks


edit -

and since you said


gravitational force = m * a

what is the value of m? which mass is it? thanks

You wanted acceleration, not force, so you divided the mass out of the equation. So, instead of
Fg= GMe/r^2 ?

it should be:

ag = GMe/r^2

which is what you were trying to find.

 

[.NoStyle]

ok thanks everyone for your help. I appreciate it.

 

[dynamicsolo]

I want to mention something further about finding the satellite's acceleration. You used the form of the centripetal acceleration equation involving the angular frequency of the orbit to get

a_c = (2·pi/21600 sec)^2 · 1.676·10^7 m = 1.42 m/(sec^2) .

You could also have used the version involving the linear velocity on the orbit:

v = C/T = (2·pi·r)/T or sqrt[G·(Me)/r] = 4875 m/sec , and

a_c = (v^2)/r ,

which is otherwise equivalent to the form you used.

You could also use the gravitational force equation, to find

F_g = G·Me·m/(r^2) = F_c = m·a_c ,

m being the satellite's mass [this equation, BTW, also gives us the "weight" of the spacecraft at this altitude above the Earth's surface -- more on this presently] , which leads to

a_c = G·Me/(r^2)
= [6.67·10^-11 N·(m^2)/(kg^2)] · (5.98·10^24 kg) / [(1.676·10^7 m)^2]
= 1.42 m/(sec^2) .

Since m does not appear in this result, this shows that any small satellite experiences the same acceleration on this orbit.

There is yet another approach using comparison ratios. The gravitational force between Earth and the satellite obeys the same relation at any radius from the Earth's surface outward. So the acceleration the satellite would experience at the Earth's surface would be

a' = G · Me / (R^2) ,

where R is the Earth's radius. You may recognize this expression as giving g , the acceleration of gravity at the Earth's surface. We can now compare this to the acceleration in the satellite's orbit:

a_c / g = [ G·Me/(r^2) ] / [ G·Me/(R^2) ] = (R/r)^2 .

In this method, we don't even need to know the value of G or the Earth's mass (much less the satellite's mass, m), since the constants all cancel out in the comparison. So we find

a_c = [ (R/r)^2 ] · g = [ (6378 km / 16,758 km)^2 ] · (9.81 m/sec^2)

= (0.3806^2) · (9.81 m/sec^2) = 1.42 m/(sec^2) .

This is another way of saying that the satellite in orbit weighs 0.3806^2 = 0.145 (about 1/7) of what it weighs on Earth.

Many roads leading to the same place (because they're all really basically equivalent)...

 

[.NoStyle]

Wow, thanks a lot for explaining that clearly dynamicsolo, I appreciate it! That's helpful to know. :)