Homework Help: TO BOBPURSLEY

Posted by TO BOBPURSLEY on Saturday, September 12, 2009 at 5:03pm.

acceleartiononmoon= GMassmoon/radius^2
=G 7.35E22/(1.74E6)^2= G 2.43E10=
= 6.67E-11*2.43E10=
1.62 m/s^2 or as a gravitational field constant= 1.62 N/kg

I have no idea why you used the mass of Earth.

were did you get

a = r^-2 Gm

from

i thought that the equation was

net force = m a = Fg = r^-2 G m m2

TO BOBPURSLEY - bobpursley, Saturday, September 12, 2009 at 5:09pm
Goodness.

forcegravity= GMassmoon*massyou/r^2
forcegravity/massyou = GMassmoon/r^2
"g"= G*Massmoon/r^2
TO BOBPURSLEY - QUESTION, Saturday, September 12, 2009 at 5:23pm
ok net force = m a

only force present

force of gravity

Fg = mg = r^-2 G m M

were m is the mass of the moon and the M is the mass of what ever it's orbiting...

correct?

Fg = mg = r^-2 G m M

divide both sides by the mass of the moon

g = r^-2 G M

so I guess I just don't get it

why would we say that the force of gravity often called weight of the moon is the mass M (what ever mass the moon is orbiting) times acceleration

force of gravity acting on the moon is equal to the Mg that's the only way we can cancel out the second mass but why would we say that the force of gravity acting on the moon is equal to the mass of the moon times gravity...
TO BOBPURSLEY - bobpursley, Saturday, September 12, 2009 at 5:28pm
you dont get it.

Divide the force of gravity by the mass of the object which is being accelerated.
F/massyou = your acceleration
of course, forces come in couplets, so
F/massmoon= moons acceleration toward you.

where F is the force of gravity, it operates on you and the moon, opposite directions.
TO BOBPURSLEY - QUESTION, Saturday, September 12, 2009 at 5:49pm
Fg exerted by whatever it's orbiting onto moon = r^-2 G (mass of moon) M = to the gravitional force exerted by moon onto M = Mg

ok then you can divide by M

but

why is it the force of gravity exerted by the moon onto whatever it's orbiting mass M just equal to Mg ???

Please I think I almost got it
TO BOBPURSLEY - QUESTION, Saturday, September 12, 2009 at 6:41pm
ok i get it see new question

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