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 with out 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


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


3. The attempt at a solution
So for some reason I bet that equation is NOT correct. Ive 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 dont understand why they would even give me the mass of the satelite if im 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:

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 dont 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

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. im so lost.

teggenspiller

what you're saying makes sence, i just dont 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.