You still might want to know basic facts about each of the planets... just saying...
Solar System B
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brobo
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Re: Solar System B
But its still a good idea to know the formula for Escape Velocity: (G2M/R)^0.5
You still might want to know basic facts about each of the planets... just saying...
You still might want to know basic facts about each of the planets... just saying...
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waawamellon
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Re: Solar System B
uhhh, teach me about 'Escape Velocity' formula because I've never even heard of it.robodude wrote:But its still a good idea to know the formula for Escape Velocity: (G2M/R)^0.5
You still might want to know basic facts about each of the planets... just saying...
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amerikestrel
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Re: Solar System B
It's basically the velocity that an object need to have to escape from the gravitational field of another object (such as a planet). So, if a satellite is orbiting a planet, it will break away from it's orbit if it reaches it's escape velocity.waawamellon wrote:uhhh, teach me about 'Escape Velocity' formula because I've never even heard of it.robodude wrote:But its still a good idea to know the formula for Escape Velocity: (G2M/R)^0.5
You still might want to know basic facts about each of the planets... just saying...
help
I don't know the formula for it though, so I can't really help you on that.
Not competing in the 2011-12 season.
2011 B division PA State Ornithology and Fossils champ!
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2011 B division PA State Ornithology and Fossils champ!
Medal count: 14
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brobo
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Re: Solar System B
Escape velocity is the velocity something must be traveling to escape the gravity of a planet. So if I wanted to throw a base ball so hard and fast that it escaped the planet, then I would have to throw it so that the velocity was equal or grater than the escape velocity of the Earth. The formula is only if something is traveling in a straight line, however:
(m2g/r)^0.5
The power to .5 is the same as square root, just FYI.
The square root of m times g time 2, divided by the r, where g= the gravitional constant (6.67*10^-11), r = radius of the planet, and m = planet's mass.
I hope that helps!
(m2g/r)^0.5
The power to .5 is the same as square root, just FYI.
The square root of m times g time 2, divided by the r, where g= the gravitional constant (6.67*10^-11), r = radius of the planet, and m = planet's mass.
I hope that helps!
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waawamellon
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Re: Solar System B
actualy, that DID help! thank you!!!!!!!!
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amerikestrel
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Re: Solar System B
Wait, I think I might of gotten something wrong.
If an object escapes the gravitational field of a planet, did it have to reach the escape velocity of the planet or it's own escape velocity?
If an object escapes the gravitational field of a planet, did it have to reach the escape velocity of the planet or it's own escape velocity?
Not competing in the 2011-12 season.
2011 B division PA State Ornithology and Fossils champ!
Medal count: 14
2011 B division PA State Ornithology and Fossils champ!
Medal count: 14
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brobo
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Re: Solar System B
No. Each planet has its own velocity. It doesn't matter whether you are throwing a baseball, a couch, or a giant building (somehow), they all have to reach the same velocity. Now, you would have to work a lot harder to throw couch than a baseball, but it still has to reach the same velocity. Remeber, size and mass does NOT the gravitational pull on that item! A baseball would fall just as fast as a bowling ball.
Does it make sense now?
Does it make sense now?
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andrewwski
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Re: Solar System B
Yes and no. For all objects of a reasonable size, their mass is negligible when compared to the mass of say, the Earth, so yes, the gravitational pull isn't really determined by the object's mass.robodude wrote:Remeber, size and mass does NOT the gravitational pull on that item! A baseball would fall just as fast as a bowling ball.
Does it make sense now?
If you're talking something like the moon on the other hand, it's going to be different, because it has a significant mass.
Newton's Law of Universal Gravitation states that:
Where
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Flavorflav
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Re: Solar System B
This is correct for force, but not for escape velocity, which is derived by setting the initial kinetic energy equal to the initial gravitational potential energyandrewwski wrote:Yes and no. For all objects of a reasonable size, their mass is negligible when compared to the mass of say, the Earth, so yes, the gravitational pull isn't really determined by the object's mass.robodude wrote:Remeber, size and mass does NOT the gravitational pull on that item! A baseball would fall just as fast as a bowling ball.
Does it make sense now?
If you're talking something like the moon on the other hand, it's going to be different, because it has a significant mass.
Newton's Law of Universal Gravitation states that:
Whereis the force of gravity, G is the universal gravitational constant (
),
and
are the masses of the two objects, and r is the distance between their centers of mass.
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brobo
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Re: Solar System B
Ok, so we are all in agreence that to calculate escape velocity, you use
where M is the mass of the planet, G is the Gravitational constant, and R is the radius of the planet.
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