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# Orbit speed Watch

1. Here's a pic of the question.

And this is the answer given

I understand what they did to find the force acting on satellites 2 and 3.

It's from the line after the v1=v2 -> T1=T2 that I don't understand what they're doing...
2. bump
3. (Original post by purplerainn)
Here's a pic of the question.

I understand what they did to find the force acting on satellites 2 and 3.

It's from the line after the v1=v2 -> T1=T2 that I don't understand what they're doing...
I wouldn't bother trying to understand what "they" did as their method is unnecessarily convoluted. Who was "they"?

The forces question is done by stating that for the satellite (S2) with mass 2m at the same orbital radius as S1, the force is simply 2 x the force on the satellite with mass m.
This is because the force is GMm/(2R)2 in the one and GM(2m)/(2R)2 in the other.
There is absolutely no need for all that calculation. Just 2 x 10,000N

For the outer satellite S3 with mass m the force F3 is proportional to 1/r2 as previously pointed out, (Newton's Law) so taking the force on S1 as 10000N the force on S3 is by proportion using inverse square

F3/F1 = (2R)2/(3R)2

F3 = 10000 x (2/3)2 = 4444N

The orbital period of S2 is of course the same as S1 because it is at the same distance from Earth.
The orbital period of S3 is found from Kepler's Law

(Period)2 is proportional to (orbital radius)3

So, using the period given for S1 and the radius 2R, and the radius of S3 as 3R you get

(T3)2 / (T1)2 = (R3)3 / (R1)3

(T3)2 = (T1)2 x (3R)3 / (2R)3

Hence (T3)2 = 2502 x (3/2)3

There is absolutely no need for all that nonsense in the answer you have been given.

My only comment would be to add that this is only a part question, clearly, and there is always a danger when answering questions such as yours where the whole question and the context has not been provided.
You may, for example, have had to derive Kepler's Law before using it.
There is no way I can know this from what you have provided.
However, a question such as this can and should be answered using simple ratios.
4. (Original post by Stonebridge)
I wouldn't bother trying to understand what "they" did as their method is unnecessarily convoluted. Who was "they"?

The forces question is done by stating that for the satellite (S2) with mass 2m at the same orbital radius as S1, the force is simply 2 x the force on the satellite with mass m.
This is because the force is GMm/(2R)2 in the one and GM(2m)/(2R)2 in the other.
There is absolutely no need for all that calculation. Just 2 x 10,000N

For the outer satellite S3 with mass m the force F3 is proportional to 1/r2 as previously pointed out, (Newton's Law) so taking the force on S1 as 10000N the force on S3 is by proportion using inverse square

F3/F1 = (2R)2/(3R)2

F3 = 10000 x (2/3)2 = 4444N

The orbital period of S2 is of course the same as S1 because it is at the same distance from Earth.
The orbital period of S3 is found from Kepler's Law

(Period)2 is proportional to (orbital radius)3

So, using the period given for S1 and the radius 2R, and the radius of S3 as 3R you get

(T3)2 / (T1)2 = (R3)3 / (R1)3

(T3)2 = (T1)2 x (3R)3 / (2R)3

Hence (T3)2 = 2502 x (3/2)3

There is absolutely no need for all that nonsense in the answer you have been given.

My only comment would be to add that this is only a part question, clearly, and there is always a danger when answering questions such as yours where the whole question and the context has not been provided.
You may, for example, have had to derive Kepler's Law before using it.
There is no way I can know this from what you have provided.
However, a question such as this can and should be answered using simple ratios.
Thanks very much.

I wasn't aware of Kelper's Law. It's a general worksheet covering various topics from the same unit; topics that have not been covered in class were also included.

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Updated: April 2, 2013
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