Gravitation and Newton's Synthesis

[pointintime]

Homework Statement

At what distance from teh Earth will a spacecraft on the way to the Moon esperience zero net force due to these two bodies because the EArth and Mon pull with equal and opposite forces?

Homework Equations

net force = m a

a radial = r^-1 v^2

Fg = r^-2 G M m

The Attempt at a Solution

for this prolem I'll use M for Moon and m for Earth

net force radial = mass of spacecraft times radial acceleration = 0 = Fg exerted by Earth = Fg exerted by moon

(mass spacecraft ) r^-1 V^2 = Fg exerted by earth

(mass spacecraft ) r^-1 4 pi^2 r^2 t^-2 = r^-2 G M m
(mass spacecraft ) r 4 pi^2 t^-2 = r^-2 G M m
r^-3 = (mass of spacecraft ) 4 pi^2 t^-2 G^-1 M^-1 m^-1
r = ( (mass of spacecraft ) 4 pi^2 t^-2 G^-1 M^-1 m^-1 )^(-3^-1)

not sure how to do this problem sense it's apparent i did it wrong
could you walk me through it

 

[RoyalCat]

There's no circular motion to speak of here. You're asked solely about equilibrium.

The moon exerts a gravitational force on the spacecraft , as does the earth.

I assume that these are all known constants: m_{earth} m_{moon}, d_{earth-moon} and the mass of the spacecraft is m

Just look at the spacecraft , what forces act on it? They're both gravitational pulls!

F_g=-\frac{GMm}{r^2}\hat r where r is the distance between the point masses in question (Masses with spherical symmetry are equivalent to point masses, the dimensions of the spacecraft are negligible when compared with the distances involved, so it can be treated as a point mass as well)

 

[Doc Al]

I suspect that they want you to ignore the motions of Earth and moon. Just set the gravitational forces equal. Hint: If the distance from the Earth's center is X, what's the distance from the moon's center?

Edit: RoyalCat beat me to it!

 

[pointintime]

i'm still not sure how to do this
there would be two varialbes the distance to the moon
and the distance to earth...

 

[D H]

The distance between the Earth and Moon is known. There is but one variable.

 

[pointintime]

so... then if I know the distance to the moon from Earth how do I know the distance from the satalite to the moon or the distance from the Earth to the satalite

 

[D H]

The Moon is some known distance from the Earth. The point in question, is some unknown distance from the Earth and is between the Earth and the Moon. Call the distance between the Earth and Moon R and the distance between the Earth and the point in question x. How far is it from the point to the Moon? Draw a picture.

 

[pointintime]

I'm not sure is it just halfway?

 

[D H]

Don't guess. You don't know where it is (yet), so use a symbol to represent the location of the point.

 

[pointintime]

http://img6.imageshack.us/img6/7346/asdfasdfasdfadsfasdf.jpg
still don't see how to do this because there are two unkown varialbes

radius of satalite to earth
Res

radius of satalite to moon
Rms

don't see how to solve with two unknowns

 

[Doc Al]

As has been stated, there's only one unknown. Look up the distance between Earth and Moon.

 

[pointintime]

I don't see how know the radius from Earth to moon is about 3.84 E 8 meters helps

 

[Doc Al]

I don't see how know the radius from Earth to moon is about 3.84 E 8 meters helps

The two distances you need must add up to the total distance from Earth to moon. Thus you end up with only one unknown.

 

[pointintime]

ok then mass m1 is earth
mass m2 is moon
mass satalite is m3
radius satalite to Earth is Rm1
radius satalite to moon is Rm2

net force acting on m3 on radial direction = m3 (acceleration in radial direction) = 0 = Force of gravity exerted on m3 by m1 = force of gravity exerted on m3 by m2

= m3 a = Rm1^-1 G m1 m3 = Rm2^-1 G m2 m3 = 0

a = Rm1^-1 G m1 = Rm2^-1 G m2 = 0

were Rm1 + Rm2 = radius moon to Earth = 3.84 E 8 m
m1 = mass of Earth = 5.98 E 24 kg
m2 = mass of moon = 7.36 E 22 kg

not sure were to go from there

 

[pointintime]

a = Rm1^-1 G m1 = (3.84 E 8 m - Rm1 )^-1 G m2 = 0
I need help rearanging for Rm1

i got his

Rm1 = (m2 - 1)^-1 (3.84 E 8 m) m1

 

[D H]

From post #1:

Fg = r^-2 G M m

That is correct. Unfortunately, that is not what you used in posts 15 and 16.

 

[pointintime]

how are you suppose to do this
how come the mass of the satalite dosen't cancel out

 

[D H]

You used gravity as a 1/r, rather than 1/r², relation in posts #15 and 16.

 

[pgardn]

Maybe a conceptul apporach before the math?And then do the math.

If the only two bodies present were the Earth and the satellite, the Earth would pull on the satellite and the satellite would pull on the earth... yes? And the force would actually be equal, the Earth would pull as hard on the satellite as the satellite pulls on the Earth, yes?

If you were able to increase the mass of the Earth (keeping the satellite the same distance away from the earth) the Earth would pull harder and the satellite would pull harder right back... If you kept the Earth the same size, but moved the satellite further away, you could calculate this force also, and it would be smaller than the two examples above. You have equations that can calculate all of these forces. You just have to know the mass of both objects, and the distance from the center of each to the other. Yes?

So now you add the moon into the situation, it is smaller than the earth, so it does not pull as hard as the Earth on a satellite halfway between the Earth and the moon, yes?

So would you not have to move the satellite closer to the moon to get it to pull just as hard as the bigger Earth? Well how much closer to the moon would the satellite have to be in order that the Earth pulled on the satellite with the same force as the moon?

You have two forces that are exactly opposite directions both pulling on the satellite, one from the big earth, and one from the small moon, so the satellite must be closer to the moon. How much closer? (Dont worry about how hard the satellite pulls on the Earth and the moon)

Now you have equations to calculate the pull on the satellite by the Earth and the pull on the satellite from the moon. Where to put the satellite so that the "pulls" would be the same but in opposite directions... kind of cancelling each other out?


EARTH_______________________________<-----sat---->____________moon

 

[pointintime]

m3 = mass of satalite

(G m3 m2)/x^2 = (G m3 m1)/(3.84 E 8 m - x)^2

I can cancel out the m3 and the G right?

 

[pointintime]

m2 = masss of Earth
m1 = mass of moon

 

[pgardn]

m3 = mass of satalite

(G m3 m2)/x^2 = (G m3 m1)/(3.84 E 8 m - x)^2

I can cancel out the m3 and the G right?

yep

is the little m meters?
take care with the inverse squares underneath when solving for x

sorry for entering the thread if I am not of help...

 

[pointintime]

ok so I got down to this

x = sqrt( ((m1 + m2)(3.84 E 8))/m2 )

which gave me the wrong answer

 

[pointintime]

yes m is meters

 

[D H]

Where does the square root come from? Show you work.

 

[pgardn]

ok so I got down to this

x = sqrt( ((m1 + m2)(3.84 E 8))/m2 )

which gave me the wrong answer

Sorry outside.


I get a quadratic...

 

[pointintime]

now what

x + m1^-2 x m2^2 = m1^-2 (3.84 E 8 m) m2^2

 

[pgardn]

Do the math your way.

I get Me((d-x)^2) = Mm(x^2)

Where d is the distance from the Earth to the moon, Me is the mass of the earth, Mm is the mass of the moon. You solve for x... your way I guess. This has now become a math problem. I hope the physics is clear.

 

[pointintime]

i guess i need some assistance with da math

x^-2 m1 = (3.84 E 8 m - x)^-2 m2