Weight on a Hypothetical Planet

AI Thread Summary
The discussion revolves around calculating a person's weight on a hypothetical planet with a mass of 1/100 that of Earth and a radius of 1/4 that of Earth. The initial attempt to derive the weight on the new planet resulted in an incorrect value due to a misunderstanding of the ratios in the gravitational force equation. Participants clarified that the mass of the person can be determined using Earth's gravitational force and that the ratios of mass and radius were mistakenly applied. The correct approach involves using the known gravitational constant and the ratios without needing the actual values of Earth's mass and radius. The conversation emphasizes the importance of careful calculations and unit consistency in solving physics problems.
raptik
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Homework Statement


The mass of a hypothetical planet is 1/100 that of Earth and its radius is 1/4 that of Earth. If a person weight 600 N on Earth, what would he weigh on this planet

Homework Equations


F = (G x M x m)/(R2)

The Attempt at a Solution


Well I know the mass of the person is the same so I rearrange this equation in terms of m and make it equal to the two equations based on their different conditions:
m = (Fe x Re2)/(Me x G) = (Fp x Rp2)/(Mp x G)

Working out for Fp I get Fp = (Mp/Me) x (G/G) x (Re2/Rp2) x Fe = 100 x 1 x (1/16) x 600N = 3750 N.

This is wrong. Somebody please help me understand what I'm doing wrong. What's the error in my method?
 
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You have the equation for force exerted by the planet.

F = (G x M x m)/(R2)

You know G (a constant), M (1/100 the mass of the earth), and R (1/4 the radius of the earth).

All you need is the mass of the person. If the Earth exerts 600N on this person, then his mass is equal to 600N/(9.8N/Kg). Just plug in all your values and it should be correct. I think you are just overcomplicating the process of finding the mass of the person. And also make sure that all of your units are correct.

If you think about it, the answer you have doesn't make sense. The force should me much smaller since the mass of the planet is much smaller while the radius doesn't change that much.
 
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guitarguy1 said:
Just plug in all your values and it should be correct. I think you are just overcomplicating the process of finding the mass of the person. And also make sure that all of your units are correct.

If you think about it, the answer you have doesn't make sense. The force should me much smaller since the mass of the planet is much smaller while the radius doesn't change that much.

So the only way to solve this is to know the actual values of Earth's mass and Radius and adjust to their respective ratios and plug in the m? I was hoping there would be a way to simply utilize the ratios without having to know the Earth's mass or Radius. Also, I understand that my answer is wrong, but could you tell me where exactly I'm going off course. In theory, I think my idea works but clearly it doesn't. Could somebody explain why?
 
raptik said:
Working out for Fp I get Fp = (Mp/Me) x (G/G) x (Re2/Rp2) x Fe = 100 x 1 x (1/16) x 600N = 3750 N.

Your method is fine. You just got the ratios Mp/Me and Re2/Rp2 backwards when you plugged in the numbers.
 
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raptik said:
So the only way to solve this is to know the actual values of Earth's mass and Radius and adjust to their respective ratios and plug in the m? I was hoping there would be a way to simply utilize the ratios without having to know the Earth's mass or Radius. Also, I understand that my answer is wrong, but could you tell me where exactly I'm going off course. In theory, I think my idea works but clearly it doesn't. Could somebody explain why?

Like I said before, to find the mass you don't need to use the mass and radius of the earth. The gravitational force on the surface of the Earth is 9.8 N/Kg. So take 600N/(9.8N/Kg) and it will give the mass of the person.
 
Doc Al said:
Your method is fine. You just got the ratios Mp/Me and Re2/Rp2 backwards when you plugged in the numbers.

Oh! Clearly. I can't afford to have those kinds of stupid mistakes on an exam. Thnx for pointing that out, I would have totally overlooked that.
 
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