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Calculating mass of sun with a graph of Logr against LogT

I have been set a task where I need to find the mass of the sun using a log graph and the equation T^2=((4pi^2)/GM))xr^3. I know that I would need to plot logT against logr where T = orbital period and r = distance from sun but what do I do with the gradient of this graph?
(edited 2 years ago)
Reply 1
Original post by Helpstudent402
I have been set a task where I need to find the mass of the sun using a log graph and the equation T^2=((4pi^2)/GM))xr^3. I know that I would need to plot logT against logr where T = orbital period and r = distance from sun but what do I do with the gradient of this graph?


You could do it that way, but it's not what I would recommend.

Your equation is T2 = (4π2/GM)r3. Note that the part in the bracket is a constant. Compare this to the equation of a straight line, y = mx + c. In your case, since you have a bit of knowledge about what this equation describes, you would expect it to go through the origin (a planet at a distance of nothing from its Sun would take no time to orbit!), so we should actually be comparing it to y = mx. Can you see what to plot on the vertical ("y") axis and what to plot on the horizontal ("x") axis?
Original post by Pangol
You could do it that way, but it's not what I would recommend.

Your equation is T2 = (4π2/GM)r3. Note that the part in the bracket is a constant. Compare this to the equation of a straight line, y = mx + c. In your case, since you have a bit of knowledge about what this equation describes, you would expect it to go through the origin (a planet at a distance of nothing from its Sun would take no time to orbit!), so we should actually be comparing it to y = mx. Can you see what to plot on the vertical ("y") axis and what to plot on the horizontal ("x") axis?

I am still not entirely sure, are u suggesting that I plot T^3 on y axis and r^2 on x axis?
Reply 3
Original post by Helpstudent402
I am still not entirely sure, are u suggesting that I plot T^3 on y axis and r^2 on x axis?

You've got the powers the wrong way around, but yes, that's what I would do. Can you see what the gradient of your graph would be if you did that?
Original post by Pangol
You've got the powers the wrong way around, but yes, that's what I would do. Can you see what the gradient of your graph would be if you did that?

The only thing with that is, I get some crazy high values which result in me not knowing how to scale my axes.
e.g my T^2 values are :
5.78x10^13
3.77x10^14
9.96x10^14
3.52x10^15
1.40x10^17
8.62x10^17
and similar issue for r^3
1.91x10^32
1.27x10^33
3.35x10^33
1.18x10^34
4.72x10^35
2.95x10^36
Reply 5
Original post by Helpstudent402
The only thing with that is, I get some crazy high values which result in me not knowing how to scale my axes.
e.g my T^2 values are :
5.78x10^13
3.77x10^14
9.96x10^14
3.52x10^15
1.40x10^17
8.62x10^17
and similar issue for r^3
1.91x10^32
1.27x10^33
3.35x10^33
1.18x10^34
4.72x10^35
2.95x10^36

Hmm. That would seem to be a practical problem, despite this being the best theoretical way to do it. Would the log values be any better though? Worth checking what they are before diving in to that alternative way of doing it.
Original post by Pangol
Hmm. That would seem to be a practical problem, despite this being the best theoretical way to do it. Would the log values be any better though? Worth checking what they are before diving in to that alternative way of doing it.

the log values don't really produce a proper trend of some sort so, I don't think they do. How would I find M if I did use logs, do you know?
Log r:
10.7627
11.0342
11.7149
11.3577
11.8913
12.4583
Log T
6.88100
7.28812
7.49904
6.37268
7.77347
8.57310
8.96780
Reply 7
Original post by Helpstudent402
the log values don't really produce a proper trend of some sort so, I don't think they do. How would I find M if I did use logs, do you know?
Log r:
10.7627
11.0342
11.7149
11.3577
11.8913
12.4583
Log T
6.88100
7.28812
7.49904
6.37268
7.77347
8.57310
8.96780

Ah - the fact that those values could be reasonably plotted on an axis is presumably the reason why you might use logs in this case. Unfortunately, it's a bit messier to do the theory. Start by taking logs of both sides of the original equation, and then use logarithm laws to break it up a bit. You want to aim for something of the form a.log(t) = b.log(r) + something else, where a and b are constants.
Original post by Pangol
Ah - the fact that those values could be reasonably plotted on an axis is presumably the reason why you might use logs in this case. Unfortunately, it's a bit messier to do the theory. Start by taking logs of both sides of the original equation, and then use logarithm laws to break it up a bit. You want to aim for something of the form a.log(t) = b.log(r) + something else, where a and b are constants.

would it be 2LogT=3logr + log4pi^2 - logGM
Reply 9
Original post by Helpstudent402
would it be 2LogT=3logr + log4pi^2 - logGM

Yes, but I would leave the last two terms as a single logarithm. It would make it easier to compare to y = mx + c, and you should then be able to see what to plot on each axis.
Original post by Pangol
Yes, but I would leave the last two terms as a single logarithm. It would make it easier to compare to y = mx + c, and you should then be able to see what to plot on each axis.

I would plot 2LogT on y axis and 3logr on x axis?
Reply 11
Original post by Helpstudent402
I would plot 2LogT on y axis and 3logr on x axis?

I think that's the best way to do it. How will you find M?
Original post by Pangol
I think that's the best way to do it. How will you find M?

will the gradient of the graph = the ((4pi^2)/GM) and I would rearrange for M but idk how to do that when I have got the log infront of that bit.
Reply 13
Original post by Helpstudent402
will the gradient of the graph = the ((4pi^2)/GM) and I would rearrange for M but idk how to do that when I have got the log infront of that bit.

It's not going to be anything to do with the gradient. Think about the +c part.
Original post by Pangol
It's not going to be anything to do with the gradient. Think about the +c part.

The y intercept of the graph will be equal to log((4pi^2)/GM)?
Reply 15
Original post by Helpstudent402
The y intercept of the graph will be equal to log((4pi^2)/GM)?

Yep, and now you have the answer to your original question. Sorry to have taken you round the houses with my first suggestion, I hope you can see that it's theoretically simpler, but it wasn't really practical with the values you have to use.
How would I solve for M tho in the log? Also thank you for being patient with me

Original post by Pangol
Yep, and now you have the answer to your original question. Sorry to have taken you round the houses with my first suggestion, I hope you can see that it's theoretically simpler, but it wasn't really practical with the values you have to use.
Reply 17
Original post by Helpstudent402
How would I solve for M tho in the log? Also thank you for being patient with me

Do you know the inverse operation for log? For example, if log(x) = 3, do you know how to find x?
Original post by Pangol
Do you know the inverse operation for log? For example, if log(x) = 3, do you know how to find x?

10^3=x
x= 1000?
Reply 19
Original post by Helpstudent402
10^3=x
x= 1000?

So you did 10 to the power of both sides. That's what you have to do here.

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