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Radius and Weight and the Earth

  • Thread starter susan__t
  • Start date
20
0
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
 

Answers and Replies

dynamicsolo
Homework Helper
1,648
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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 .
 
20
0
Thank you that's wonderful, I ended up with 24000 Newtons which sounds about right
 
dynamicsolo
Homework Helper
1,648
4
That looks right, but it sounds painful... (That's almost 5400 lbs.!)
 
DaveC426913
Gold Member
18,327
1,940
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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