Determining distance between geostationary satellite & moon

[dakota224]

***NOTE: The data in my written work & the typed variables under "all known data" are correct. The problem statement has new values for a new submission, but I am trying to correct my work on an old submission. Sorry for the confusion!

1. Homework Statement
Problem:

All known data:
Period (T) = 402 days = 34732800 seconds
Moon's Diameter = 4900 km = 4900000 m
Moon's Radius = D/2 = 2450000 m
Density Moon = 4500 kg/m³
Volume of Moon = (4/3)pi(2450000m)^2 = 61600872 * 10¹⁹ m³

Homework Equations

r³ = (T²Gm)/(4pi²)
r = ((T²Gm)/(4pi²)) ^ (1/3)

The Attempt at a Solution

I may have just made a math mistake, but I ran through this twice and arrived at the same result. The only thing I can think of that I may still need to do after arriving at 824250792m is somehow find the radius of the Death Star then subtract that? According to this explanation of the formula I used (http://www.dummies.com/how-to/content/how-to-calculate-the-period-and-orbiting-radius-of.html) that would not be necessary.

 

[SteamKing]

Homework Statement

Problem:
View attachment 96272
All known data:
Period (T) = 402 days = 34732800 seconds
Moon's Diameter = 4900 km = 4900000 m
Moon's Radius = D/2 = 2450000 m
Density Moon = 4500 kg/m³
Volume of Moon = (4/3)pi(2450000m)^2 = 61600872 * 10¹⁹ m³

Homework Equations

r³ = (T²Gm)/(4pi²)
r = ((T²Gm)/(4pi²)) ^ (1/3)

The Attempt at a Solution

View attachment 96273

I may have just made a math mistake, but I ran through this twice and arrived at the same result. The only thing I can think of that I may still need to do after arriving at 824250792m is somehow find the radius of the Death Star then subtract that? According to this explanation of the formula I used (http://www.dummies.com/how-to/content/how-to-calculate-the-period-and-orbiting-radius-of.html) that would not be necessary.

The density of the moon is 3500 kg/m³ according to the problem statement. You have used a density of 4500 kg/m³ for some reason.

Always check your work for silly mistakes like this. It might save points on exams and assignments.

 

[dakota224]

The density of the moon is 3500 kg/m³ according to the problem statement. You have used a density of 4500 kg/m³ for some reason.

Always check your work for silly mistakes like this. It might save points on exams and assignments.

Oops, that's not a work mistake, just a typo I forgot to check. The online program that gives the homework changes the data/values after each submission, so the problem statement has a new value of 3500, but my work is for 4500. I'm trying to correct my work, then I can start over with the new values once I figure out the correct steps to the solution.

 

[SteamKing]

Oops, that's not a work mistake, just a typo I forgot to check. The online program that gives the homework changes the data/values after each submission, so the problem statement has a new value of 3500, but my work is for 4500. I'm trying to correct my work, then I can start over with the new values once I figure out the correct steps to the solution.

Why do you think your original result was in error?

 

[dakota224]

Why do you think your original result was in error?

I know it was incorrect through the submission software. Other than a math error, looking over it, the best guess I have as to why it's wrong is that I need to do one more step of subtraction. I'd need to check for other formulas to see if I have enough information, but if I could solve for diameter of the Death Star, I could then subtract the radius of it from my previous answer to get the exact distance between the moon and Death Star...

 

[SteamKing]

I know it was incorrect through the submission software. Other than a math error, looking over it, the best guess I have as to why it's wrong is that I need to do one more step of subtraction. I'd need to check for other formulas to see if I have enough information, but if I could solve for diameter of the Death Star, I could then subtract the radius of it from my previous answer to get the exact distance between the moon and Death Star...

From what I understand of these things, submissions sometimes are marked wrong if they don't have the correct number of significant figures.

Good luck on finding information about the Death Star. Remember, it's fake.

 

[dakota224]

From what I understand of these things, submissions sometimes are marked wrong if they don't have the correct number of significant figures.

Good luck on finding information about the Death Star. Remember, it's fake.

That is possible, but before my first submission it said "tolerances for answers are quite large" - plus, since data in the problem changes with each submission, testing different answers by changing significant figures isn't possible.

As for the Death Star, I didnt mean "look up" information about the fictional object :)

I meant, using data given and data I found, could I solve for the diameter of it?

 

[SteamKing]

That is possible, but before my first submission it said "tolerances for answers are quite large" - plus, since data in the problem changes with each submission, testing different answers by changing significant figures isn't possible.

You'll have to decide on the proper number of sig figs while working out the solution then.

Since these calculations involve G, that's one of the key values which limit the precision of the final result.

As for the Death Star, I didnt mean "look up" information about the fictional object :)

I meant, using data given and data I found, could I solve for the diameter of it?

I don't see how. You don't know its density or diameter or its mass.

 

[dakota224]

You'll have to decide on the proper number of sig figs while working out the solution then.

Since these calculations involve G, that's one of the key values which limit the precision of the final result.

I don't see how. You don't know its density or diameter or its mass.

I really do not think the answer was denied due to a rounding error. I have done 30+ assignments on this platform and getting the right answer, but not submitting it with the correct number of sig figs has never been an issue. I am looking for a review of my work/formulas...is how I solved for the answer correct in terms of the physics? Can I solve for a geostationary satellite, moving around a tidal-locked satellite moon of another planet, using the method I did?

 

[SteamKing]

I really do not think the answer was denied due to a rounding error. I have done 30+ assignments on this platform and getting the right answer, but not submitting it with the correct number of sig figs has never been an issue. I am looking for a review of my work/formulas...is how I solved for the answer correct in terms of the physics? Can I solve for a geostationary satellite, moving around a tidal-locked satellite moon of another planet, using the method I did?

Well, you can always repeat your method for a geostat orbiting the earth. The answer to that is well known.

 

[SteamKing]

I really do not think the answer was denied due to a rounding error.

To be clear here, rounding error is one thing, sig figs is another separate issue.

Your calculations showed an orbital radius of 826700792 m, which is 826,701 km, rounded to the nearest kilometer.

However, G = 6.67 × 10^-11 is known only to 3 significant figures, which would make your orbital radius 827,000 km.

The question asks "How far is the Death Star from the moon?" That's a tad ambiguous.

The only reasonable interpretation, IMO, with the given data, is to give the center-center distance between the two bodies.

 

[dakota224]

Well, you can always repeat your method for a geostat orbiting the earth. The answer to that is well known.

Does that mean the planet is meaningless towards solving the problem? If you can

To be clear here, rounding error is one thing, sig figs is another separate issue.

Your calculations showed an orbital radius of 826700792 m, which is 826,701 km, rounded to the nearest kilometer.

However, G = 6.67 × 10^-11 is known only to 3 significant figures, which would make your orbital radius 827,000 km.

The question asks "How far is the Death Star from the moon?" That's a tad ambiguous.

The only reasonable interpretation, IMO, with the given data, is to give the center-center distance between the two bodies.

Ok, I see how those numbers adjusted to significant figures would have a pretty significant impact on the result. I will try again, making sure I'm consistent with rounding and sig figs, and find just the center to center distance without subtracting out the radius of the moon. I'll let you know if that works.

 

[SteamKing]

Does that mean the planet is meaningless towards solving the problem?

It's not clear what you mean here.

If by 'planet' you mean the body about which the geostat orbits, no, it's not meaningless. For the earth, you want the orbital period of the geostat to be exactly 23h56m and you can look up the mass and diameter of the earth.

 

[dakota224]

I tried again with consistent rounding to 2 decimal places and followed a different method getting there, still the incorrect answer:

 

[dakota224]

I tried again with consistent rounding to 2 decimal places and followed a different method getting there, still the incorrect answer:View attachment 96298

Also, I did not include this work on the above screenshot, but I also did it with the "old" equation r = cube root ((T²Gm)/(4pi²)) and got the same result - 39743 m. What am I doing wrong? I must have the wrong understanding of the entire problem. I do not understand how the Planet Endor is relevant to the problem, as these equations are simply treating the Death Star as a satellite of the moon...but that is a tidal-locked satellite on its own.

 

[SteamKing]

Also, I did not include this work on the above screenshot, but I also did it with the "old" equation r = cube root ((T²Gm)/(4pi²)) and got the same result - 39743 m. What am I doing wrong? I must have the wrong understanding of the entire problem. I do not understand how the Planet Endor is relevant to the problem, as these equations are simply treating the Death Star as a satellite of the moon...but that is a tidal-locked satellite on its own.

One mistake I quickly noticed was that you calculated the wrong volume for the moon.

For a sphere, V = (4/3)π ⋅ R³, but you have used V = (4/3)π ⋅ R

 

[gneill]

Were you given any information about Endor?

If the Death Star's orbit is geostationary above the moon, that means the same location on the moon must always be below it. But the moon is tidally locked with Endor. So what does a "circular geostationary orbit" really mean? Can you draw such a scenario including all three bodies?

 

[dakota224]

One mistake I quickly noticed was that you calculated the wrong volume for the moon.

For a sphere, V = (4/3)π ⋅ R³, but you have used V = (4/3)π ⋅ R

Wow, cannot believe I did that. That was the problem - answer accepted. Thanks for all your help!