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# The gravity on a rocket planet three times the size of the earth...?

...would be what, and would it be tolerable to lifeforms from earth?

Update:

Ooops, I mean "rocky planet"

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• 1 month ago

Depends on the density of the planet.

The further out from the center of gravity of the planet you are, the less you weigh;  however, the *more mass* there is, the *more* you'd weigh.

So... Earth has a density of 5.5... Let's assume the volume of this new planet is 3 times that of the Earth, and maintains the same density - so, the mass of the new planet is also 3 times that of the Earth.

V = 4/3 x pi x r^3...

For Earth, our radius is "1" - so:

V = 4/3(3.141592)(1)^3

V = 4.189.

With the new planet, 3V = 4/3 x pi x r^3... so:

3(4.189) = 4/3 x (3.141592) x r^3

which boils down to

3 = r^3, so r = 1.442

The formula for finding the force of gravity between two objects is:

F = G(m1)(m2) / r^2, where the force F equals the gravitational constant G, times the mass of the first object (the planet) times the mass of the second object (say it's you or me), all divided by r^2.

So, the force of gravity on the new planet is going to increase the mass of M1 by 3 times, and it's going increase the radius by 44.2%:

F = G(3m1)(m2) / (1.442^2)   (since G and m2 don't change, let's remove them.)

F = 3/1.442^2 = 3/2.079 =  1.445

so - if you weigh 100 newtons on Earth, you'd weigh 144.5 newtons on the new planet.

• Anonymous
1 month ago

What exactly is a "rocket planet".

"3 times the size of the Earth" is meaningless.  3 times the mass, 3 times the diameter, both?

Surface gravity is a function of mass and radius.  We need both to determine surface gravity.

And if life evolved in that environment it would survive just fine.

• 1 month ago

The size of the planet is not what is important. It's the MASS of the planet that determines  the magnitude of the force of gravity. 3 times the Mass would be?3 times the gravity. Most people would find that intolerable.