- #1
Balti
- 6
- 1
- Homework Statement
- Two masses of ##9.0×10^{24}kg## and ##5.0×10^{24}kg## respectively, are separated by ##1.6×10^{11}##m. A third mass is placed between the other two masses, such that it remains stationary. How far from the ##9.0×10^{24}kg## mass is the third mass placed? Give your answer in giga-metres (Gm) but without a unit. Give your answer to two significant figures.
- Relevant Equations
- ##g=\frac {F}{m}
Morning all
I've recently come across a problem where I get conceptually but cannot apply mathematically if that makes sense.
I understand the position of the third mass must be at the equilibrium point of ##m_1## (##9.0×10^{24}kg##), so ##\Sigma F = 0## right? And not even necessarily zero, something must equal zero for the mass to be at the neutral point or the Lagrange point.
The farthest I've come is having something like ##\frac {9}{x^2} = \frac {5}{(1.6×10^11-x)^2}## and even then I have no idea how I got to that.
Can someone please give me any tips? I don't necessarily want answers, but just hints or a clue.
I've recently come across a problem where I get conceptually but cannot apply mathematically if that makes sense.
I understand the position of the third mass must be at the equilibrium point of ##m_1## (##9.0×10^{24}kg##), so ##\Sigma F = 0## right? And not even necessarily zero, something must equal zero for the mass to be at the neutral point or the Lagrange point.
The farthest I've come is having something like ##\frac {9}{x^2} = \frac {5}{(1.6×10^11-x)^2}## and even then I have no idea how I got to that.
Can someone please give me any tips? I don't necessarily want answers, but just hints or a clue.