# Math Help - gravitational force

1. ## gravitational force

need help with this please. not sure if there is an equation for gravitational force, and which one to use.

The drawing shows one alignment of the sun, earth, and moon.
The gravitational force SM that the sun exerts on the moon is perpendicular to the force EM that the earth exerts on the moon.
mass of sun = 1.99 1030 kg,
mass of earth = 5.98 1024 kg,
mass of moon = 7.35 1022 kg.

The distances shown in the drawing are rSM = 1.50 1011 m and rEM = 3.85 108 m. Determine the magnitude of the net gravitational force on the moon.

in Newtons

What is rSM and rEM?

need some help with the equations please.

heres whats going on: http://img91.imageshack.us/img91/2585/p424cw9.gif

2. I reckon you'll be able to do this one if you put ur mind to it.

For this question, we need to find the force of attraction between the Earth and the Moon, and also the force of attraction between the Sun and the Moon. From there we can find the total force on the moon.

It will help a lot if you draw a quick diagram of the information you know for this one. You should basically have a triangle with the Sun, Moon and Earth at each corner.

We know the mass of each and the distance between the Eart and the Moon, so plug that info into the formula and get that force.
Remember the formula F = (M1*M2)/(R^2)?

We know the mass of the sun and the moon, but not the distance separating the, so let's find that first.
From your diagram, we can find a triangle with sides 385,108m and 1,501,011m. Using Pythagoras' Theorem we can find the hypotenuse.
Now we have the distance between the Sun and the Moon and the mass of each, so we can apply the formula to find the force of attraction.

Now we have 2 forces acting on the moon. By adding these vectors, we can find the total force acting on the moon.
Let me know if you're not familiar with vectors and I'll explain more.

3. okay, i guess i need a further explanation, because i'm still not understanding parts of this.

i thought the distance between the sun and moon was known: The distances shown in the drawing are rSM = 1.50 x 10^11 m

also, why do i need the hypotenuse for this problem?

thankyou.

4. Originally Posted by Spimon
I reckon you'll be able to do this one if you put ur mind to it.

For this question, we need to find the force of attraction between the Earth and the Moon, and also the force of attraction between the Sun and the Moon. From there we can find the total force on the moon.

It will help a lot if you draw a quick diagram of the information you know for this one. You should basically have a triangle with the Sun, Moon and Earth at each corner.

We know the mass of each and the distance between the Eart and the Moon, so plug that info into the formula and get that force.
Remember the formula F = (M1*M2)/(R^2)?

We know the mass of the sun and the moon, but not the distance separating the, so let's find that first.
From your diagram, we can find a triangle with sides 385,108m and 1,501,011m. Using Pythagoras' Theorem we can find the hypotenuse.
Now we have the distance between the Sun and the Moon and the mass of each, so we can apply the formula to find the force of attraction.

Now we have 2 forces acting on the moon. By adding these vectors, we can find the total force acting on the moon.
Let me know if you're not familiar with vectors and I'll explain more.

I thought the equation wa : F = G(M1*M2)/(R^2)

I am having problems with this one too.

5. Originally Posted by rcmango
okay, i guess i need a further explanation, because i'm still not understanding parts of this.

i thought the distance between the sun and moon was known: The distances shown in the drawing are rSM = 1.50 x 10^11 m

also, why do i need the hypotenuse for this problem?

thankyou.
Originally Posted by La_Diabla
I thought the equation wa : F = G(M1*M2)/(R^2)

I am having problems with this one too.
First off, spimon made a simple typo. The formula for Newtonian gravity is indeed
$F = \frac{Gm_1m_2}{r^2}$

So let's define a coordinate system with +x horizontally to the left and +y upward.

The force between the Earth and the Moon will be
$F_{ME} = \frac{GM_EM_M}{r_{EM}^2}$
and since this is an attraction it will be in the negative y direction.

The force between the Sun and the Moon will be
$F_{MS} = \frac{GM_SM_M}{r_{SM}^2}$
and since this is an attraction it will be in the positive x direction.

Now you need to add these forces vectorally.

-Dan