Alright guys, here's a really hard one I cannot get.
A newly-discovered planet has only one moon. Their center-to-center distance is 790 000 000 metres. The planet's mass is 90 times the mass of its moon. How far from the center of the planet would a space shuttle have to be so it would experience no net force?
This is all the info I got.
Any help?
Possible equations?
Gravitational force between two objects is explained as
F_g=G\frac{m_1m_2}{r^2}
where m is the mass of the two objects, G is the gravitational constant (6.673*10^-11) and r is center to center distance between the two objects.
We then want the shuttle to be in a position where the force exerted by the moon and the planet are equal! (thus net force is zero).
let the following be true:
mm = mass of the moon
mp = mass of the planet
ms = mass of the ship
fm = force between moon and ship
fp = force between planet and ship
as stated above, the forces must be equal, therefore:
F_m=F_p
G\frac{m_mm_s}{r_1^2}=G\frac{m_pm_s}{r_2^2}
Now if we think about r here, if the ship is between who objects that are only 100m away, we say that if it is 10m from object a, then it must be 90m from object be. Or if the distance equals x from a, then the distance must be 100-x from b. so here we can write that the distance the ship is from the moon be r, and the distance from the planet be 7.9*10^8-r. Therefore we have:
G\frac{m_mm_s}{r^2}=G\frac{m_pm_s}{(7.9*10^8-r)^2}
Lets start cancelling!
\frac{m_m}{r^2}=\frac{m_p}{(7.9*10^8-r)^2}
But wait! were also given that the mass of the planet is 90 times the moon, so we can say:
m_p=90*m_m
we then have:
\frac{m_m}{r^2}=\frac{90*m_m}{(7.9*10^8-r)^2}
Factor the bottom out and bit more cancellation with mm
\frac{1}{r^2}=\frac{90}{(7.9*10^8)^2-2r(7.9*10^8)+r^2)}
Some algebra work:
(7.9*10^8)^2-2r(7.9*10^8)+r^2)=90r^2
Combine like terms!
0=89r^2+2r(7.9*10^8)-(7.9*10^8)^2
This becomes just a quadratic equation :cool:, solve for r
