1. ## satellite orbit problem

A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 4.80 x 10^4 s. What is the speed at which the satellite travels?

should be in m/s

not sure about what the equation is and how to reach the speed.

2. Originally Posted by rcmango
A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 4.80 x 10^4 s. What is the speed at which the satellite travels?

should be in m/s

not sure about what the equation is and how to reach the speed.
It's moving in a circle right? So answer these two questions:
1) What is the net force on the satellite?
2) What force(s) is/are causing this net force?

-Dan

3. Originally Posted by rcmango
A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 4.80 x 10^4 s. What is the speed at which the satellite travels?

should be in m/s

not sure about what the equation is and how to reach the speed.
$F_{g}=F_{c}$

$\frac{Gm_{1}m_{2}}{r^2}=\frac{m_{1}v^2}{r}$

$\frac{Gm_{2}}{r}=v^2$

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

$\frac{Gm_{2}}{r}=\frac{(2\pi r)^2}{T^2}$

$Gm_{2}T^2=(2\pi r)^2 r$

$\frac{Gm_{2}T^2}{(2\pi )^2}=r^3$

$\sqrt [3] {\frac{Gm_{2}T^2}{(2\pi )^2}}=r$

$G=6.67*10^{-11}$
$m_{2}=5.98*10^{24}$
$T=4.8*10^4$
$r=28552879$

$g=Gm_{2}/r^2$

$g=.489245914923$

$critical velocity = \sqrt {rg}$

$critical velocity=3737.56m/s$

It think.
There's probably a shorter way

4. Originally Posted by hummeth
sq root of rg=critical velocity
What critical velocity?
$F_G = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$
where m is the mass of the satellite, M is the mass of the Earth, and r is the distance from the center of mass of the satellite to the center of mass of the Earth. (r is not simply the height above the ground.)

So
$GM = rv^2$

Now,
$v = \frac{2 \pi r}{T}$
where T is the period. (Note that this is true only for circular motion.)

Inserting this equation into our "GM" equation:
$GM = r \left ( \frac{2 \pi r}{T} \right ) ^2$

$GM = r \cdot \frac{4 \pi ^2 r^2}{T^2}$

$GM = \frac{4 \pi ^2 r^3}{T^2}$ <-- Note: We could have started from here using one of Kepler's Laws. (The 2nd?)

Thus
$r = \sqrt[3]{ \frac{GMT^2}{4 \pi ^2}}$

Up until now, this is identical to hummeth's solution.

Now recall the GM equation:
$GM = rv^2$

$GM = v^2 \sqrt[3]{ \frac{GMT^2}{4 \pi ^2}}$

So finally:
$v = \sqrt{ \frac{GM}{\sqrt[3]{ \frac{GMT^2}{4 \pi ^2}}}}$

We can simplify this a bit, but as the result gives nothing of value, you can simply calculate it from here. I get that v = 3737.56 m/s.

-Dan

5. I would say the force involved is the centripetal force of weight pushing toward the earth.

I'm still not sure how this answer was reached.

other than some formulas like Fc = (m * v^2) / r

and then: sqrt( (G * Me) / r )

and then for then T = period,
so, v = (2 * pi * r) / T

i still could not reach the correct answer because variables were missing like r

still need help with this one.

6. plug in G as $6.67*10^-11$
Mass and Period are given
$r=28552879.9$
$
v=2\pi r/T
$

plug in r T and you get 3737.56

7. Originally Posted by hummeth

plug in G as $6.67*10^-11$
Mass and Period are given
$r=28552879.9$
$
v=2\pi r/T
$

plug in r T and you get 3737.56
I must have miscalculated. I agree with hummeth's answer.

-Dan

8. I miscalculated too. My original and second way both work, if you do the arithmetic correctly!

9. Originally Posted by hummeth
I miscalculated too. My original and second way both work, if you do the arithmetic correctly!
Yes but my question to you, if I were grading this, is where did you get the equation $r = \sqrt[3] { \frac{GMT^2}{4 \pi ^2} }$ from? You can get it from one of Kepler's Laws (I still haven't looked up which), but you should at least mention that.

Equations "pulled out of the hat" can be dangerous. Use your powers wisely.

-Dan

10. Didn't we both get it from $F_{g}=F_{c}$?

11. Originally Posted by hummeth
Didn't we both get it from $F_{g}=F_{c}$?