Help with physics problem

[frostedflakes]

Is from the test I just took, problem has me completely stumped. Our teacher is great at throwing curve balls at us on the tests, you study a lot and think you know what you're doing but he still manages to put something completely different on there that you haven't seen before.

Using the following information, determine the amount of time in years it takes for Mars to orbit the sun.

The average distance from the sun to Mars is 1.52 times the average distance from the sun to Earth.
The mass of Mars is 0.11 times the mass of Earth.
The average radius of Mars is 0.53 times the average radius of Earth.

What we know:
R(mars)=1.52R(earth)
M(mars)=0.11M(earth)
r(mars)=0.53r(earth)
G=6.67*10^-11

The tools:
T^2=(4*pi^2*R^3)/(GM)
F=(GMm)/(R^2)

The answer:
???

It's easy enough to solve if you know the mass of the sun or radii of the orbits, I'm just stumped on how it can be solved without this information. Any ideas?

 

[TuxDave]

You have the fact that the Earth takes 1 year to rotate the sun and how the Mass and Radius is relative to Earth. If you half the radius and keep the mass the same, how much faster must the angular velocity be... etc....

 

[rbV5]

Originally posted by: frostedflakes
Is from the test I just took, problem has me completely stumped. Our teacher is great at throwing curve balls at us on the tests, you study a lot and think you know what you're doing but he still manages to put something completely different on there that you haven't seen before.

Using the following information, determine the amount of time in years it takes for Mars to orbit the sun.

The average distance from the sun to Mars is 1.52 times the average distance from the sun to Earth.
The mass of Mars is 0.11 times the mass of Earth.
The average radius of Mars is 0.53 times the average radius of Earth.

What we know:
R(mars)=1.52R(earth)
M(mars)=0.11M(earth)
r(mars)=0.53r(earth)
G=6.67*10^-11

The tools:
T^2=(4*pi^2*R^3)/(GM)
F=(GMm)/(R^2)

The answer:
???

It's easy enough to solve if you know the mass of the sun or radii of the orbits, I'm just stumped on how it can be solved without this information. Any ideas?

It takes exactly 1 martian year.

 

[frostedflakes]

I'm sorry man, I'm still not getting it.

What equation(s) can be used to derive the relationship between radius of orbit and period? EDIT: Wow brain fart, law of periods gives this relationship. Sorry, stupid question.

EDIT: Lol rbV5, I thought about putting that down, but it probably would have been guaranteed 0 partial credit. At least I have a chance of getting some partial credit by trying to work the problem.

 

[silverpig]

T(earth)^2 ~ R(earth)^3

T(mars)^2 ~ R(mars)^3

T(mars) / T(earth) = [R(mars)^3 / R(earth)^3]^(1/2)

T(earth) = 1

So you just have:

T(mars) = [R(mars)^3 / R(earth)^3]^(1/2)

Then sub in R(mars)=1.52R(earth)

T(mars) = [{1.52R(earth)/R(earth)}^3]^(1/2)

T(mars) = [ 1.52^(3/2) ]

T(mars) = 1.87 earth years

Which is, I believe, the correct answer.


(all that pi^2, G, M stuff cancels out)

 

[frostedflakes]

Ah I think I get it now, don't know how this threw me off so bad, I think I was trying to make it more complicated than it should have been.

Thanks for all the help silverpig and TuxDave (and rbV5 )

 

[silverpig]

Originally posted by: TuxDave
You have the fact that the Earth takes 1 year to rotate the sun and how the Mass and Radius is relative to Earth. If you half the radius and keep the mass the same, how much faster must the angular velocity be... etc....

It's completely independent of mass.