Beginner Physics: Solving G Constant on Planets

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To find the gravitational constant (g) on a planet with the same density as Earth but double the radius, the equation Fg = Gm1m2/d^2 can be used. The relationship between mass and volume indicates that if the density remains constant, the mass of the planet will increase with the cube of the radius. Therefore, with a radius of 2r, the mass becomes eight times that of Earth. Plugging this mass into the equation simplifies to g = G(8m)/(2r)^2, which ultimately leads to g being four times that of Earth's gravitational constant. This approach clarifies the calculations needed to solve the problem effectively.
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Hi, everyone I'm new to physics and I got a problem that I can't solve.
It says
"What would the g constant be on a planet that has the same density as Earth but has twice the radius?"

Can someone help me on this. Also can someone tell me if I'm suppose to use the equation: Fg=Gm1m2/d(square)?
 
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I'm new to physics too (so don't take my word for this, lol), but my guess is you set that equation above equal to ma, or in this case mg...

so mg = G m1 m2/ r^2, then the masses of the object on the planet cancel out so g = G m (planet)/ (2r)^2

since the radius is twice that of Earth its just 2r..then plug in the values of G and r, and I'm not sure about mass of planet. if the densities are the same, are the masses are the same? if they are, then just plug in the mass of earth...and calculate g

hope this helps a bit until someone smarter comes along, lol
 
Thanks for the tips, now i think I'm getting some of it.
 
PhysicBeginner said:
Hi, everyone I'm new to physics and I got a problem that I can't solve.
It says
"What would the g constant be on a planet that has the same density as Earth but has twice the radius?"
Can someone help me on this. Also can someone tell me if I'm suppose to use the equation: Fg=Gm1m2/d(square)?

If its looking for a numerical answer, I'd find the average density of Earth, find the radius, double the radius, then find the amount of mass that fits in that radius, and put all our new data for the imaginary Earth into the equation you cited at the end.
 
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