Determine Orbital Period Of Satellite

[Bashyboy]

Homework Statement

An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.32 m/s^2. Determine the orbital period of the satellite [in minutes].

Homework Equations

$g= G \frac{M_E}{r^2}$ Solving for $r$, and G is a constant

$T^2=K_sr^3$ Solving for T, and $K_s$ is a constant.

The Attempt at a Solution

Solving the first equation for $r$, I obtain the value of $6.30440 \cdot 10^{13}~m$; and substituting this value into the second equation, I get an orbital period of $1.24 \cdot 10^{21} min$ (this value has been converted from seconds to minutes). However, this isn't the correct. What did I do incorrectly?

 

[SteamKing]

What value of Ks are you using?

 

[Bashyboy]

That would be the one they provided in the textbook, 2.97 x 10^-19.

 

[haruspex]

Solving the first equation for $r$, I obtain the value of $6.30440 \cdot 10^{13}~m$

Did you stop to think how far that is? Looks like you forgot to sqrt.

 

[Bashyboy]

I re-worked the problem, but it is still incorrect.

I recalculated the distance at which the satellite is suspended above the earth:

$r= \sqrt{\frac{G(5.97 \cdot 10^{24})}{6.32}} = 7940022.399~m$

Using this, I evaluated the period equation: $T= \sqrt{K_s(7940022.399)^3} = 12.2~s$

None of this is correct. What am I doing wrong? I am not sure if these are the proper equations.

 

[Dick]

I re-worked the problem, but it is still incorrect.

I recalculated the distance at which the satellite is suspended above the earth:

$r= \sqrt{\frac{G(5.97 \cdot 10^{24})}{6.32}} = 7940022.399~m$

Using this, I evaluated the period equation: $T= \sqrt{K_s(7940022.399)^3} = 12.2~s$

None of this is correct. What am I doing wrong? I am not sure if these are the proper equations.

You probably don't want to use the Kepler equation. The acceleration due to gravity should be the same as the centripetal acceleration v^2/r. Can you figure out a way to use that?

 

[Chestermiller]

You don't even need to know the mass of the Earth to do this. All you need to know is that g varies inversely as the square of the distance from the center of the Earth and that the radius of the Earth is about 6450 km. So

$$(9.8)(6450)^2=6.32r^2$$

What does this give you for r?

 

[haruspex]

Yes, simply adjusting from g by the ratio of squares of the radii is much the simplest, but in terms of where the OP went wrong it was in using K_s, which incorporates the mass of the sun, instead of (what I assume may be called) K_e.

 

[Chestermiller]

Yes, simply adjusting from g by the ratio of squares of the radii is much the simplest, but in terms of where the OP went wrong it was in using K_s, which incorporates the mass of the sun, instead of (what I assume may be called) K_e.

As I'm sure you're aware, in the next step, he didn't need to know the mass of the Earth either. He could just simply have used:
$$\omega^2r=6.32$$
with $\omega=\frac{2\pi}{T}$

Chet

 

[Bashyboy]

All right, I tried it a different way, according to the suggestion of Dick.

I still used the same method to find the radius at which the satellite is revolving.
That being: $r=7940022.399~m$.

Then, $6.32=\frac{v^2}{7940022.399} \implies v = 7083.85~m/s$

Furthermore, $v=r\omega \implies \omega = 8.92 \cdot 10^{-4}~rad/s$

Using the relationship Chestermiller alluded to, $\omega = \frac{2\pi}{T} \implies T=0.001401476~s$

However, after all is said and done, this is the incorrect answer.

 

[Chestermiller]

All right, I tried it a different way, according to the suggestion of Dick.

I still used the same method to find the radius at which the satellite is revolving.
That being: $r=7940022.399~m$.

Then, $6.32=\frac{v^2}{7940022.399} \implies v = 7083.85~m/s$

Furthermore, $v=r\omega \implies \omega = 8.92 \cdot 10^{-4}~rad/s$

Using the relationship Chestermiller alluded to, $\omega = \frac{2\pi}{T} \implies T=0.001401476~s$

However, after all is said and done, this is the incorrect answer.

You messed up the algebra in the very last step. Otherwise, OK.