Radius and Weight and the Earth

AI Thread Summary
To determine Pat's weight on a planet with twice Earth's mass and one quarter of its radius, the gravitational force formula Fg = G(m1)(m2)/r² is used. By setting up a comparison ratio of weights on both planets, the exact values for Earth's mass and radius are not needed. The weight on the other planet can be calculated using the known weight on Earth (750 N) and the ratios of mass and radius. The result indicates that Pat would weigh 24,000 Newtons on this hypothetical planet, which raises concerns about the physical plausibility of such conditions. Overall, the discussion emphasizes the importance of understanding gravitational ratios in physics problems.
susan__t
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The question my teacher gave me is as follows...
If Pat weighs 750N on Earth, what would Pat weigh on another planet with twice Earth's mass and one quarter the Earth's radius?
We are not given the exact radius or mass of the Earth and I am having a hard time figuring it out.
I know I have to used the formula Fg =G(m1)(m2)/r squared
I considered also trying to find some kind of ratio (like 2/(.25)squared) but I don't understand how that will help me and what I have to do from there

My test is tomorrow, please help!
Thanks,
Susan
 
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susan__t said:
I know I have to used the formula Fg =G(m1)(m2)/r squared
I considered also trying to find some kind of ratio (like 2/(.25)squared)

Yes, you want to set up a comparison ratio; that way, you don't need to know the exact mass and radius of Earth.

Call the weight of the person on Earth W_e . That is given by

W_e = G·(mass of person)·(mass of Earth) / (radius of Earth)^2 .

Call their weight on the other planet W_p . How would you write that?

Since you want to compare the two, what is the quotient (W_p)/(W_e)? That is to say, if you take the ratio of the two expressions, what cancels out and what remains?

You are given the ratio of the planet's radius to Earth's radius and the ratio of the planet's mass to Earth's mass. You will get a numerical value for (W_p)/(W_e) . Since you know the person's weight on Earth (W_e = 750 N), you can now find their weight on the other planet, W_p .
 
Thank you that's wonderful, I ended up with 24000 Newtons which sounds about right
 
That looks right, but it sounds painful... (That's almost 5400 lbs.!)
 
dynamicsolo said:
That looks right, but it sounds painful... (That's almost 5400 lbs.!)
On a planet that's 30x denser than Earth...You wonder if the teacher could have come up with a scenario that could at least physically exist.
 
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