Need help with gravitational field question

Try writing down the equation for gravitational field strength twice - once using the information for the Earth and once using the information for the Moon. You should then be able to combine then into one equation and re-arrange it so that (Re/Rm) is on one side - this should give you a numerical value for the ratio

Thanks for the reply - I understand that one now.

I don't understand how to do most of these ratio questions though, there's nothing in my book to say how to generally go about them. These two questions have completely stumped me again, I feel like I've tried everything lol. Could you get me started on them? They're from the multiple choice section of AQA June 2013.

13) Two stars of mass M and 4M are at a distance d between their centres. The resultant gravitational field strength is zero along the line between their centres at adistance y from the centre of the star of mass M. What is the value of the ratio?
A – 1/2
B – 1/3
C – 2/3
D – 3/4

14) Mars has a diameter approximately 0. 5 that of the Earth, and a mass of 0. 1 that of theEarth. The gravitational potential at the Earth’s surface is –63 MJ kg–1. What is the approximate value of the gravitational potential at the surface of Mars?
A –13 MJ kg–1
B –25 MJ kg–1
C –95 MJ kg–1
D –320 MJ kg–1

The second one I ended up arriving at B (which was incorrect, it's actually A) by writing that -63x10^6 = -GM/r, and then I multiplied both sides by 0.1 and divided by 0.25, since Mars' mass was 0.1 of the Earths and its radius was a quarter.

Thanks!

Firstly, you are close to getting the answer to the second one, but you've been incredibly silly. If I tell you we have a circle of diameter 10m. I then tell you circle two has a diameter half of that of that first, i.e 5m, what's circle two's radius? Remember the radius of the first is 5m. Compare the two radii.

For 13, we know the resultant gravitational field strength is zero at this point.

Write down two equations for the strength at this point due to either mass, and equate them.

I.e:

$\dfrac{GM}{y^2} = \dfrac{4GM}{(d-y)^2}$

Cancel this down, then factorise, and finding $\dfrac{y}{d}$ should be trivial.

Thanks, but how the hell do you arrive at y/d from that equation?

I cancel GM from both sides, bring the (d-y)^2 and the y^2 up to get (d-y)^2 = 4y^2 and then square root both sides to get d-y = 2y but then how do i get rid of the -y? I think I need to brush up on my maths skills!