Question about gravity and planets

  • Thread starter Thread starter jakeowens
  • Start date Start date
  • Tags Tags
    Gravity Planets
AI Thread Summary
The discussion centers on the relationship between a planet's mass, diameter, and the resulting weight experienced by an individual. It is established that weight is determined by the gravitational force, which depends on both mass and the radius of the planet. A planet with four times the mass of Earth and two times its diameter would result in a weight increase of 16 times, due to the squared relationship of radius in the gravitational formula. The diameter significantly impacts gravitational strength, confirming that both mass and size are crucial in calculating weight. Ultimately, understanding these principles clarifies how weight varies across different planetary bodies.
jakeowens
Messages
34
Reaction score
0
If your on a planet that has 4x as much mass as earth, but only 2x the diameter, would you weigh 4x as much as you do on earth? Does the diameter of the planet even matter? or is it only the mass that affects how much you weigh.

Thanks
 
Physics news on Phys.org
hmm, i'll try,

weight is w=mg, where g is the gravitational field strength at a point. Weight is the G force acting on you.

using F= G \frac{Mm}{r^2} where M is mass of Earth and r is radius of earth, according to your example,

F= G \frac{4Mm}{(\frac{1}{2r})^2}

notice that half r is being squared, this gives you a quarter r ! If you bring everything up, and simplify,

you get,

F= 16 G \frac{Mm}{r^2}

so it weighs 16 times more on earth.

The diameter of the planet matters, because the equation depends on r.
 
Last edited:
misogynisticfeminist said:
hmm, i'll try,

weight is w=mg, where g is the gravitational field strength at a point. Weight is the G force acting on you.

using F= G \frac{Mm}{r^2} where M is mass of Earth and r is radius of earth, according to your example,

F= G \frac{4Mm}{(\frac{1}{2r})^2}

I think this equation should be:

F= G \frac{4Mm}{(2r)^2}

The 4 and the 22 will cancel out and you'll end up with 1g.
 
just sitck some values into it and see how it varies. pick F=forget about constant F=Mm/d^2 and replace everytihng you don't need by 1 and everytihng you need by 2 and 4 accordingly. work smart...work simple.

F=4Mm/(2D)^2 so the 4 cancels the 2^2.
 
ohhh thought the radius was half.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top