Recent content by tiffany chan

Got it! Its 20R! Thank you very much~

If the force = 0 when the GP is maximised. That means, gravitational field strength at that point = 0 as well. So, the equation will be: 100GM/x^2 - GM/(22R-x)^2 = 0, then solve for x?

When potential is maximised, the gravitational force should be zero?

I use x because i considered it as ratio. So correct way would be (-100GM/X) + (-GM/(22R-x)) = total GP? But I’m confuse of what value to be put for the total GP in the equation

1. Assume the maximum distance happened at a distance x from the centre of the planet. 2. Total Gravitational Potential would be: (-100GM/x) - (-GM/1-x) = (what should be the max GP equal to)??

Thanks for your reply. So do you mean -GM/R is not the right equation to be used to solve this question?

By using the equation for the Gravitational Potential -GM/R. It is understand that the max Gravitation Potential would be at infinity point. - G(100M)/22R-(-GM/22R) would be the maximum Gravitational Potential... I guess... Then what would be the next step to find the distance?