Universe Sandbox
General Category => Astronomy & Science => Topic started by: atomic7732 on March 22, 2011, 07:35:25 PM
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6.673*10^-11 N m^2/kg^2*((mass of earth*mass of jupiter)/81362282064194900 m^2)
I didn't do this right...
So a Jupiter mass planet at the distance of Sirius will exert a force of 9.30323757 × 10^24 newtons on earth?
How would I calculate how much this affects Earth's orbit? Would I need to factor in the sun?
The sun exerts a force of 5.28599732 × 10^34 m^3 kg / s^2? WTF??? It's not even in newtons...
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I think this is correct:
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wtf
and how did you type it like all mathematical?
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What do you mean by wtf?
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Did you pay money for this?
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He did.
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Yeah, I got the demo.
Sucks it aint a calculator. Then I'd pay for it. :P
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Bla I think you did it just fine except for one error... It should be N(m^2/kg) not N(m/kg)^2, as shown in my proof that Google makes calculations much more difficult than they could be.
Anyhow the result you got (.002 N) is much too small for my theory of a new way to find extrasolar planets. With a very very very sensitive instrument, it would work, but you would need to subtract the motion of the earth cause by the planets, and their satellites (which should create more of an effect then a jupiter planet 8.6 ly away) as well as many of the nearby stars within around 500 ly. I'll toss some numbers and see what I come up with.
Actually I'm not sure what you did, I'm left with the unit N*kg... well you do have kg^2, so that'd be N... but when I converted the units myself I got N(m^2/kg), but I make stupid errors sometimes so who knows.
I just called it newtons anyway... and i got an extremely large number for some equation...
If this force was applied to earth, every kilogram would be exerted under a focre of over 21,000 newtons. I messed up somehow.
EDIT!!! Forgot to square the distance... wow... but then how would the meters be squared? I'm really confused...
This makes more sense...
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時々、数学は理解するに難しいである。
Sometimes ｍath is hard to understand.
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I didn't use Google, I just found the stuffs on Wikipedia and put it into the formula.
I didn't pay for the program, as I got a license from my high school.
http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation
Here it says N*m^2*kg^-2, which is the same as N*(m/kg)^2 and not N(m^2/kg).
http://en.wikipedia.org/wiki/Gravitational_constant
On this page it's written like I did.
Now I just realized I used a wrong mass for Earth. I also think I must've made a typo on my calculator, 10^41 should've probably been like 10^51.
Remember to put brackets around the radius, and squared after the bracket, so you square both the unit and the distance.
New calculashun:
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much smaller than my calculation... lol 2.18x10^18 is a nice number. :P
No, Bla, I said Google makes it much harder than the calculation needs to be. It says whatever I put on top of the bracket, I divide by the units in a newton to get a much simpler... unit. I guess.
When I try to get the acceleration of the objects, I divide the result by earth's mass, and I get a number in N/kg which is the same thing as m/(s)^2. How would I convert to m/s? Would I just square root the number?
Wolfram|Alpha says they are incompatible... So how do space simulators do this stuff? wait... wouldn't m/s/s be m/(s)^2?
It is! Problem solved!
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Adding some random vectors. :)