Question about Planetary orbit and satelites

[teggenspiller]

1. The pro blem statement, all variables and given/known data
the question:
9.0 × 10^3 kg satellite orbits the Earth at the distance of 2.56 × 10^7 m from Earth’s surface. What is its period?
They give me a multiple choice, but let's figure this out without it.

okay so my variables known are the mass of the satelite:
"MSat"- 9.0*10^3
d/r= 2.56 × 10^7 m
mearth: 5.9742 × 10^24 kilograms
G= 6.67*10^-11

Homework Equations

Kepler's Law (?)
T^2= 2*pi*r(^3/2)/G*Mp

The Attempt at a Solution

So for some reason I bet that equation is NOT correct. I've been searching all over google and through my notes and its all i can come up with. I'm not sure if I use the mass of the satelite or the mass of Earth in the equation. I don't understand why they would even give me the mass of the satelite if I am not supposed to use it though
so I put into my calculator: T=squrt: 2*(3.14)^2* 2.56 × 10^7 m (or is it the radius of earth+the distance given(?) / (6.67*10^-11)* 5.9742 × 10^24
or do i use the mass of satelite here?

 

[Mike Pemulis]

Your equation in #2 has a typo. It should be,

[tex]
T^2 = \frac{4 \pi^2 r^3}{GM}
[/tex]

This is Kepler's Third Law, which may help with Googling. See if you can find what you are supposed to plug in for r and M on your own. One hint:

I don't understand why they would even give me the mass of the satelite if I am not supposed to use it though

Yes, you're right, physics problems never give you extra information just to screw with you.

 

[teggenspiller]

well, when you put it like that...:)

 

[teggenspiller]

i used your equation. i used the satelite for mass. i tried the radius as the radius given AND the radius give+the radius of Earth itself, neither time did i get an answer even close

 

[teggenspiller]

& the period of a satellite does NOT depend on its mass, so why in the world would they give me its mass?

 

[Mike Pemulis]

Let's figure out the radius first. Our two options are the height itself, or the height plus the planetary radius. Which one is right?

Hint: if there was no atmosphere, satellites could orbit at ground level (if there were no mountains, people, etc to get in their way). Which option makes sense in that context?

 

[teggenspiller]

option B

 

[teggenspiller]

using your equation:
(sqrt(4 * (pi^2)) * (2.56 * (10^7))) + (((6 378.1^3) / (6.67 * (10^11))) * (5.9742 * (10^24))) = 2.32395597 × 10^24

all of my options' powers of ten are 4.
i don't see how/where I am going wrong.

Using the mass as the satellites mass

(sqrt(4 * (pi^2)) * (2.56 * (10^7))) + (((6 378.1^3) / (6.67 * (10^11))) * (9 * (10^3))) = 160 853 045 <---wrong as well.
hmm.:/

 

[Mike Pemulis]

Yup. And that's a general result; outside of a spherical body, the gravitational force from that body is the same as if all the mass was concentrated at the center. So in this kind of problem, r always means "distance from the center."

Now for the mass. Why did they give you both masses? To make you think about this question!

If you're still stuck, drop a textbook* and a pencil from the same height and see which one falls faster.

*Not a physics book. Drop a math or chemistry book, that's all they're good for. Edit: Looks like you have an order-of-operations error. Slow down and make sure you're putting the numbers into the calculator correctly.

 

[teggenspiller]

okay, okay. i see. So why arent my 'generally correct' results = any of my choices?

by the way, they hit the ground at precisely the same time.

 

[Mike Pemulis]

by the way, they hit the ground at precisely the same time.

Yes. So does motion in a gravitational field depend on the mass of the falling object?

(Also, I checked the answer and got [something] times 10⁴ seconds. So it does work out.)

 

[teggenspiller]

no it does not. just how far it is away from the mcenter, i suppose.

:) well thank you for the help

 

[Mike Pemulis]

I think you might just be making calculator errors. Can you show me what you're entering in?

Two things:

1. Make sure you are really dividing by the denominator.

2. The equation above has T^2 on the LHS. So you have to take the square root at the end to get T.

 

[teggenspiller]

i using those equations from above.

along with endless variations of it. including an equation of velocity, (v=sqrt(G*m^2)/R, as in distance from center, like you said.

then inserting value of v into T=2piR/V
which is a random equation I found in my notes. I am so lost.

 

[teggenspiller]

what you're saying makes sence, i just don't get how to apply it.

 

[Mike Pemulis]

Those two equations are actually how you derive my equation at the top. So let's look at those. It's a good idea to know how equations are related. So, we have:

F_g = GMm/r²

Where F is the force of gravity, G is the gravitational constant, M is the mass of the planet, m is the mass of the satellite, and r is the distance between the satellite and the center of the planet. This is the fundamental equation that describes gravity.

Now, imagine a particle moving in a circle. You have to exert a force to keep it moving in a circle, otherwise it flies off in a straight line. That force is,

F_circle = mv²/r

For a satellite, gravity is the force that keeps the satellite moving in a circle. So we should set the two forces equal to get,

mv²/r = GMm/r²

We can divide both sides by m to get and multiply both sides by r to get,

v² = GM/r

(Notice the mass of the satellite is gone, as expected).

Okay, now to introduce the orbital period. The circumference of a circle is:

[tex]
c = 2 \pi r
[/tex]

And speed v is distance divided by time. The time it takes to complete on circle is T, by definition. So we have,

[tex]
v = c/T = 2 \pi r/T
[/tex]

Okay, so as a first step, plug the last equation into v in the fourth equation, and show that the result is,

[tex]
T^2 = \frac{4 \pi^2 r^3}{GM}
[/tex]

Which is the equation from before.