Calculating the Mass of the Sun Using Earth's Orbit Data

[Miike012]

The Earth orbits the sun in an approximately circular
orbit of radius 1.496 * 10^11 m (about 93 million miles)
in a period of 365 days. Use these data to determine
the mass of the sun.

Solution:

F = (M_s)(M_e)(G)/r^2

I have two unknowns... F and M_s... please give suggestions because I have no clue where to start...

 

[K.S]

Well you have period T = 365days = 3.1536E7

Try using Kepler's 3rd law in which

F = -G(Ms)(Me)/r^2 = Me(v^2)/r , assuming it's a circular orbit - hence you can get rid of Me

You know that v = 2(pi)(r)/T

The rest of it should be pretty straightforward. :)

 

[technician]

You have the radius of the circle and the time for 1 revolution. It is circular motion so can you use this information to calculate the acceleration of the circular motion?

 

[Miike012]

Well you have period T = 365days = 3.1536E7

Try using Kepler's 3rd law in which

F = -G(Ms)(Me)/r^2 = Me(v^2)/r , assuming it's a circular orbit - hence you can get rid of Me

You know that v = 2(pi)(r)/T

The rest of it should be pretty straightforward. :)

Why do you have a negative value.. -G(Ms)(Me)/r^2 ?

 

[Miike012]

shouldnt it be possitive? and what do u mean I can remove Me by using keplers law...
keplers law is T = r^3/2... how will that remove Me?

 

[K.S]

Oh my bad, take away the minus sign,

In this case you can do so since both the centripetal force Me(v^2)/r and gravitational force G(Ms)(Me)/r^2 point towards the same direction - towards the center of the sun.

 

[Miike012]

Ok I got the right answer.. thank you...
One last question...
I got F = (Me)(4)(pi)^2*(r)/T^2

By keplers law can I set T = r^3/2 ?
But when I did this I didnt get the correct answer... can I not use keplers law in this situation?

 

[technician]

use the radius of the orbit and the time of 1 revolution to calculate the speed of the Earth (2.98 x 10^4 m/s)
acceleration in circular motion =v^2/r ...calculate the acceleration (5.94 x 10^-3)
acceleration = (G x M)/r^2 G = 6.7 x 10^-11
this gives mass of sun = 2 x 10^30kg

 

[K.S]

technician you are exactly right.

Okay I think it was confusing I threw in kepler's third law here - what I have shown you is a prediction of that law.

Earlier, I said that
F = G(Ms)(Me)/r^2 = Me(v^2)/r,

since the Earth can be seen as an object orbiting around the sun in a circular manner, and there exists a radial force of the same magnitude - the gravitational force, pulling the Earth towards the sun.

v = 2(pi)(r)/T

As technician have mentioned, you should use the time for 1 revolution, which is T,

Hence G(Ms)/r^2 = ((2(pi)(r)/T)^2)/r

It shows in the end that T^2 = (4pi^2/G(Ms))*r^3

OR

T = (4pi^2/G(Ms))^(1/2)*r^(3/2) , so you can make the comparison here

The thing is I didn't memorize Kepler's third law either, but I know how it works. You should too, try to learn to link different concepts to each other (in this case, circular motion and gravitation), rather than trying to memorize different formulae. Save that bio memory space for the next few years, for my professors have so kindly warned us that there are plenty more equations for us to memorize in the future.

 

[K.S]

Oh, and you're welcome. :)

I'm also a student, and have much to learn (Very likely from you all too!) - totally struggling with rotational physics sigh.

 

[technician]

KS... good comments from you. This part of physics is not easy. That is the value of forums like this...you get lots of different views and ways to tackle problems.

 

[cmb]

Why do you have a negative value.. -G(Ms)(Me)/r^2 ?

KS was right first time, no? Force is a vector and is opposite to the radius direction. If 'r' is the +ve distance away from a force attractor, then the force is -ve.

It doesn't change the 'magnitude' answer, but I think it is semantically more correct.

 

[K.S]

technician, thanks for the kind words! Yeah, maybe I'm just not used to rotational motion, my brain seems to spin along with the concept. Agreed, that's my main reason in joining the forum!

cmb, Force is indeed a vector. However, in this expression,

F = G(Ms)(Me)/r^2 = Me(v^2)/r ,

we're comparing the magnitudes of these forces. So it doesn't make sense to put the minus sign there... does it?

 

[cmb]

we're comparing the magnitudes of these forces. So it doesn't make sense to put the minus sign there... does it?

It makes sense if you write the semantically correct equation of motion, with the acceleration being -(v^2)/r.

F = -G(Ms)(Me)/r^2 = -Me(v^2)/r

makes no difference to the answer in this case, I was just looking to point out that -G(Ms)(Me)/r^2 wasn't wrong in your first statement. (It may be less obvious on other occasions, especially if you start doing some calculus on that acceleration term and likewise drop any -ve's that appear from the r^-1 index!)

 

[K.S]

Oh okay, that makes sense. Thanks!