Gravity Qn

[IntoTheNewWorld]

Gravity questions annoy me =[
Any help on this question (attached) greatly appreciated =]

[appianway]

I cbs working it out properly, but you could probably do it like this:

+ find g of mercury from the g of mars (g mars = 1/16 g mercury)
+g=GM/r^2 (we know the mass of the sun, so rearrange to find the the radius of mars' orbit)
+Mercury's orbit's radius = 1/4 of the orbit of mars (so then you've got the second part of the question done)
+g=4((pi)^2)r/T^2 (rearrange to find the value of T^2 for each one)
+Then sub the values of T^2 into your ratio

There's probably an easier way to do this, but I don't really have the time to sit down and look at it properly.

[appianway]

Whoops, misread the question.

[appianway]

R^3/T^2 = GM/4pi

So T^2/R^3 = 4pi/GM

Something like that.

[IntoTheNewWorld]

thx Appianway =], it's simpler than I thought =0

[appianway]

No problem... sorry for misreading the question at first!

[anonuser0511]

LOL i wasted my time working this out the hard way so i'll give you the easy way first

By Kepler's Third Law(rearranged) we got
T^2        4pi^2
___   =   ______
R^3          GM

sub M in as the Sun and you're sweet you got

2.2959E-19s^2m^-3
units will be seconds^2*meters^-3

what's interesting is that the ratio for this for every planet rotating around the sun is theoretically the same.

p.s. yea hardway is what appian said first*