earth, what will the value of gravity be on this planet?
2007-02-15 13:58:51 · 8 answers · asked by hydrogen_ions 1 in Science & Mathematics ➔ Astronomy & Space
it would be the same as earth's. gravity depends on mass, not size
2007-02-15 14:07:10 · answer #1 · answered by andrew b 3 · 0⤊ 2⤋
Gravitational attraction in newtons F = G m₁m₂/r² G = 6.674e-11 m³/kgs² m₁ and m₂ are the masses of the two objects in kg r is the distance in meters between their centers Double the radius, and the 1/r² becomes 1/2² or 1/4, so the force goes down by a factor of 4, as does the acceleration. (c)
2016-05-24 05:30:16 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The force of gravity on such a planet will be 1/4 of Earth's gravity.
2007-02-15 18:10:37 · answer #3 · answered by som 1 · 0⤊ 0⤋
Gravity on such a planet would be 1/4th the strength of gravity on earth (Since gravity is inversely proportional to the square of the radius of the planet, and directly proportional to the mass).... so acceleration due to gravity would be about 2.45 meters per second squared... 'hope that helps
2007-02-15 14:06:02 · answer #4 · answered by Nate K 1 · 4⤊ 0⤋
Gravity works an inverse-square law. The surface is twice as far from the center--sothe surface gravity will be 0.25 that of Earth (2.45 meters/sec2).
2007-02-15 14:42:55 · answer #5 · answered by Anonymous · 0⤊ 1⤋
mg = GMe m / R^2
This is Newton's Universal Law of Gravitation set equal to your weight. m is your mass. It divides out and leaves you with
g = GMe/R^2 Now, Let's double R to get g(on this new planet)
g(on new planet) = GMe /(2R)^2
g(on new planet) = GMe/(4R^2)
or
g(on new planet) = (1/4) GMe/R^2
But remember that GMe/R^2 = g on earth, so
g(on the new planet) = (1/4) g = (1/4)*9.8 m/s/s.
2007-02-15 14:10:31 · answer #6 · answered by Dennis H 4 · 0⤊ 0⤋
No, mass has to do with how dense or physically compact the planet is, not size.
2007-02-15 15:30:38 · answer #7 · answered by Anonymous · 0⤊ 1⤋
1g, 9.8 m/s2
2007-02-15 14:06:41 · answer #8 · answered by doom98999 3 · 0⤊ 2⤋