I need physics help with orbits!?

Question

Imagine that a new asteroid is discovered in the solar system with a circular orbit and an orbital period of 5 years.

What is the average distance of this object from the sun in earth units? Answer in units of AU.

If you could please show me how to do it, it would be greatly appreciated.

Answer 1

Ok, that one is not too bad.

the formula for getting the period is:

P= Sqrt(AU^3/Msun)

P = Period (years)
AU = Astronomical Units (93,000,000 mi or 149,000,000km )
Msun = Mass of star relative to mass of sun (this proportion is important if you want to know about an orbit around another star, or even of a moon around a planet.)

but we need to work t backward to get the AU, so:

P^2 = AU^3/Msun (Msun in this case = 1 so we can just drop it)
P^2 = AU^3

Cuberoot(P^2) = AU

to get a cube root, raise a number by (1/3) or .3333333333333

You can test these formulas with other orbits that are established...

BTW since this equation uses proportions, like AU instead of km, and since the mass of the asteroid is negligible compared to the star, you don't need to worry about the gravitational constant and really complicated math to get a very accurate answer.

*EDIT* as for those who think we shouldn't answer homework questions

I agree. Dont just give the answer. Offer help so that the person can understand the question, give hints as to where to find the answers, but don't just give them. If you write a question that looks like homework or a test question, you should expect help, not easy answers.

However, saying that a person should give up a course because they are having difficulty with a concept is plain stupid. The question is beneath your notice? That is arrogant. This person didn't ask for the answer, they asked for help understanding how to get the answer. It is not your place to "punish" people.

BTW, sorry I didn't mention Kepler's laws, I was pulling the formula from a spreadsheet and totally forgot to mention it. Kudos to the answerers who did!

And thanks to everyone who tries to help people understand science and math. Not everyone gets it right away, and helping them helps us stamp out poor education and ignorance. Keep asking questions--the only dumb question is the one you didn't ask (though there are some on here that are doozies).

Answer 2

This is a simple application of Kepler's Third Law. Kepler's Laws of Planetary Motion are:
1. Planets move in orbits that are ellipses
2. The planets move such that the line between the Sun and the Planet sweeps out the same area in the same area in the same time no matter where in the orbit.
3. The square of the period of the orbit of a planet is proportional to the mean distance from the Sun cubed: p^2=Ka^3.

If you state the period in Earth years and the distance in AU, the constant of proportionality K is 1. So for your asteroid,
5^2 = a^3
solve for a
a = 5 ^(2/3)

Answer 3

Let us forget about any laws other than
F=GMm/(r^2) -1
F=m(v^2)/r - 2
We know G, we (may) know M ( let us put that aside for a moment), we (may) know m ( let us put that aside for a moment)
Equating 1 and 2
GMm/(r^2)=m(v^2)/r
GMm/r=v^2 - i
As you can see, m can be eliminated.
The orbit is circular, so the velocity through the orbit is constant.
s=vt
s=2(pi)r
Therefore, 2(pi)r=vt
v=2(pi)r/t - ii
If we substitute (ii) in (i) we get
r^3=GM(t^2)/4(pi^2)
Going back to knowing M, we want R in AU, so if R if the orbital distance of Earth, T its orbital time, r the orbital distance of the asteroid, t its orbital time
(r^3)/(R^3)=(t^2)/(T^2)
Put R=1 (AU), t=1 (year)
r=5^(1/3)=1.7099667730342445850595511436608 (hahaha) AU

Answer 4

That's too trivial to be worth me answering (though I imagine it would make a great homework question, in fact that's about all it would be useful for). Ask your teacher (i.e. the person who set your assignment).

r^2 to t^3 is really so damn simple that if you're having problems you should probably reconsider whether continuing with your course is really wise since you aren't learning anything.

Answer 5

v = ?(GM/r) ... wherein M is the mass of the Earth and G is the gravitational consistent. Convert your 'r' in to meters = 6800000m Throw the values into the equation and there you will have it :) For the time interval, alternative the above equation into that for round movement (T = (2?r)/v) This offers: T = ?(four ?² r ³)/GM ;)

Answer 6

use the keppelers third equation, the r^2 prop to t^3..

r1^3 t1^2
----- = -----
r2^3 t2^2

put r1=1 AU, t1=1 yr, t2=5 yr... get the r2...