** Question related to Orbital speed of star, Keep getting *10^4 to high?

[pargeterw]

*Please Help* Question related to Orbital speed of star, Keep getting *10^4 to high??

Homework Statement

A star has a mass of 2.2*10³⁰kg, there is a planet of mass 4.5*10²⁷kg orbiting it at a distance of 3.1*10¹¹m (also told that it has a "period of wobble" of 9.2*10⁷s due to the orbiting of the planet)

The start is orbiting about the centre of mass of the star/planet system.

Show that the speed of the star in its circular orbit is about 50m/s

Homework Equations

v=sqrt((G*M)/r)
where G=6.67*10^-11m³/kg/s²
M=mass of orbiting object
r=radius of orbit

The Attempt at a Solution

I know M=2.2*10³⁰
I know G=6.67*10^-11
so I only need r

r is the distance from C.O.M of the Star/planet system to the star, and it follows that this would equal ((4.5*10²⁷)*(3.1*10¹¹))/((4.5*10²⁷)+(2.2*10³⁰)) = 6.33*10⁸
because <x>=(∑mixi)/(∑mi)

therefore substituting M, G, and r into the equation for v should, one would hope, give about 50m/s

it does in fact, give 48*10⁴.

this is slightly hopeful, as 48≈50, although not correct, as 10⁴ ≠ 1!

does anybody have any idea what I got wrong?

 

[tiny-tim]

welcome to pf!

hi pargeterw! welcome to pf!

(try using the X² icon just above the Reply box )

r is the C.O.M of the Star/planet, so it follows that it would be at a distance of ((4.5*10^27)*(3.1*10^11))/((4.5*10^27)+(2.2*10^30)) = 6.33*10^8

erm …

that's the distance of the c.o.m. from the star!

 

[pargeterw]

that's the distance of the c.o.m. from the star!

I know that, isn't that the distance that I want?

 

[ppzmis]

so distance of star from com as you say is r= 6.33E8.
period of orbit = 9.2E7
Circular orbit is 2.*pi. r/T = 43ms^-1

 

[tiny-tim]

no, you want the orbital radius r of the planet,

ie the distance of the planet from the c.o.m.

(though since the question says "about", i think you're supposed to pretend that r is just the distance given, between the planet and the star )

 

[pargeterw]

Why do I want the orbital distance of the planet when I'm trying to work out the orbital speed of the star?

 

[pargeterw]

@ppzmis - you seem to have the right answer, which is encouraging, how can you have used the Time "period of wobble" since surely this is caused by the orbiting of the planet around the star, and I'm trying to calculate the speed of the star about the C.O.M of the system?

EDIT - although surely 43m/s is about 40m/s not about 50m/s? so looking less encouraging now :(

 

[ppzmis]

My thought was that the time period of the wobble is related to the planet moving around the star but both star and planet must orbit around the centre of mass with the same period. Otherwise if you think about a star and planet orbiting around a point in space you could end up with the impossible situation where the star and planet are both on the same side of the centre of mass!

43 is ~50 to 1 sig fig! I reckon that with those numbers and the question as it stands that is the right answer. Check you've copied the numbers correctly. The radius of orbit of the star is 6.34E8 but that won't make much difference.

 

[pargeterw]

You're right about the Time period being relevant, but it appears that the main cause of the problem was me trying to be too clever for the question and cut a corner with the centre of mass thing. The truth is that both methods should work, and yours (stemming from my cheating ways) only produces a lower value than it should because the values in the question are not accurately fabricated. The most annoying thing is that while it was a completely valid method, it was not on the mark scheme, and would barely have got me any marks, if at all. I did however work out the answer the long way around, see below for your interest

Thanks a lot for helping though!

Will

v=(2πr)/T and a=(v²/r)

now a=F/M and F=(GMm)/R²
so a=((6.67*10^-11)*(4.5*10²⁷))/((3.1*10¹¹)²)
so a=3.12*10^-6ms^-2

if you do some re-arranging of the first two formulae,

a=(v²/r) --------> r=(v²)/a
and since v=(2πr)/T --------> r=(a*T²)/(4π²)

Substituting that into v=(2πr)/T and cancelling like crazy gives

v=(aT)/(2π)

v=((3.12*10^-6)*(9.2*10⁷))/(2π)

v= 45.68ms^-1 ≈ 50ms^-1 QED

:)

 

[pargeterw]

Also,

The radius of orbit of the star is 6.34E8 but that won't make much difference.

How did you get that? It's 6.327965525*10⁸ according to my calculator...