When I set F=(m1)a equal to F=G(m1)(m2)/r^2 i find out that the mass m2 depends on the distance between the two objects: (m1)a = G(m1)(m2)/r^2
a = G(m2)/r^2
a(r^2)/G = m2
as you see the mass m2 depends on r, the distance between the two objects. Can someone explain me what is it that I'm doing wrong?
2007-03-10 15:30:59 · 10 answers · asked by ? 2 in Science & Mathematics ➔ Physics
For a given a,m2 depands on r
2007-03-10 15:37:59 · answer #1 · answered by JAMES 4 · 0⤊ 0⤋
Think about what it is you are trying to do. Your system of two simultaneous equations is correct if the acceleration of m1 is due only to its gravitational attraction to m2. The first equation you derived lets you calculate the acceleration of m1. You notice, as did Galileo, that its acceleration is independent of its mass. The second equation you derived lets you calculate m2 if you know the acceleration of m1. It is correct, and you can use this method to calculate the mass of the earth. If that's what you were trying to do, you did nothing wrong. But you should say what you were trying to do, and why you took the steps you took.
2007-03-10 17:56:24 · answer #2 · answered by Frank N 7 · 0⤊ 0⤋
No. Space time can be imagined as the "fabric" of the universe. It's the thing that all objects move about in; the stuff where change and motion can occur. The universe is the term given to everything beyond the big bang including matter, energy and anything else we find.
2016-03-28 23:46:21 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Cex, it's exactly correct, it's TRUE that
(a r²) / G = m2 !
You see, if I know that I'm 6,373 km from the center of this earth (sea level), and a = g = 9.81 m/sec², we can actually compute the mass of the Earth this way!
Mass of Earth = (9.81 m/sec²(6,373 km)²) / 6.6742 * 10^-11 m³/kg-sec = 5.97 * 10^24 kg, to a good degree of accuracy.
2007-03-10 16:10:36 · answer #4 · answered by Scythian1950 7 · 1⤊ 0⤋
haha...u took the equation the wrong way...a=Gm2/r^2....here the independent quantities are m2 and r....G is a constant...this makes a and not m2 or r the dependant quantity
2007-03-10 15:56:26 · answer #5 · answered by lilmissy 2 · 0⤊ 0⤋
Keep in mind that, for circular motion:
a = V^2 / r
also.
I see what you've run into though,...
"a" is variable as well.
It has been a while since I've looked at this stuff, but 'will look again
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2007-03-10 15:36:34 · answer #6 · answered by RockHanger 3 · 0⤊ 0⤋
How about F=G(F)2F/r^1
2007-03-10 15:39:32 · answer #7 · answered by ron p 2 · 0⤊ 0⤋
is that even engish you are speaking or is some kind of made up launguage?
2007-03-10 15:34:14 · answer #8 · answered by Anonymous · 0⤊ 1⤋
Your doing that G** AWFUL problem...Ouch! You make my head hurt...thats what your doing wrong!!!!!!@#$%
2007-03-10 15:34:29 · answer #9 · answered by 123..WAIT! 5 · 0⤊ 1⤋
UGH.....WHAT?
2007-03-10 15:33:19 · answer #10 · answered by hysteria75 2 · 0⤊ 1⤋