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Weight on [Planet] Question

  • Thread starter davev
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Neptune has a masss 17.2 times larger than that of Earth and a radius 3.88 times larger. A person weighing 176 lb on the Earth would weigh how much on Neptune?

Earth's mass is 6x10^24; Earth's radius is 6.4x10^6. Neptune's mass is 1.032x10^26; Neptune's radius is 24832000. 176 lbs is 80 kgs.




Weight (or Force) = G[itex]\frac{Mplanet\bullet Mperson}{r2}[/itex]

G (universal) = 6.7x10^-11
Mperson for Earth = 80 kgs
Mplanet for Earth = 6x10^24 kgs
r for Earth = 6.4x10^6

Mperson for Neptune = ?
Mplanet for Neptune = 1.032x10^26
r for Neptune = 24832000



Every time I try to solve for the mass of the person on Neptune, I keep getting around 70 kgs, which is 154 lbs. The correct answer according to my online homework system is 201.084048 lbs.

First, I solve for Force on Earth. I plug all four values into the equation to get 785.15625 Newtons.

Second, I then plug the answer above into the Force value to solve for Mperson on Neptune. To isolate Mperson I multiply both sides of the equation by r^2, divide both sides of the equation by G, and divide both sides of the equation by Mplanet. The answer I get is 70.02 kgs, which is 154.05 lbs. That is not the correct answer.


I know that all I really need to do is divide 3.88^2 by 17.2, then multiply that value by 176, but I want to find out what I'm doing wrong. I don't want to use any shortcuts until I actually understand the long process.

Thanks!
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Neptune has a masss 17.2 times larger than that of Earth and a radius 3.88 times larger. A person weighing 176 lb on the Earth would weigh how much on Neptune?

Earth's mass is 6x10^24; Earth's radius is 6.4x10^6. Neptune's mass is 1.032x10^26; Neptune's radius is 24832000. 176 lbs is 80 kgs.




Weight (or Force) = G[itex]\frac{Mplanet\bullet Mperson}{r2}[/itex]

G (universal) = 6.7x10^-11
Mperson for Earth = 80 kgs
Mplanet for Earth = 6x10^24 kgs
r for Earth = 6.4x10^6

Mperson for Neptune = ?
Mplanet for Neptune = 1.032x10^26
r for Neptune = 24832000



Every time I try to solve for the mass of the person on Neptune, I keep getting around 70 kgs, which is 154 lbs. The correct answer according to my online homework system is 201.084048 lbs.

First, I solve for Force on Earth. I plug all four values into the equation to get 785.15625 Newtons.

Second, I then plug the answer above into the Force value to solve for Mperson on Neptune. To isolate Mperson I multiply both sides of the equation by r^2, divide both sides of the equation by G, and divide both sides of the equation by Mplanet. The answer I get is 70.02 kgs, which is 154.05 lbs. That is not the correct answer.


I know that all I really need to do is divide 3.88^2 by 17.2, then multiply that value by 176, but I want to find out what I'm doing wrong. I don't want to use any shortcuts until I actually understand the long process.

Thanks!
You are going the long way around. Mperson on Earth=Mperson on Neptune and it's about 80kg. You don't need to solve for a different mass on Neptune. Mass doesn't change. Only the weight changes.
 
Last edited:
  • #3
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You are going the long way around. Mperson on Earth=Mperson on Neptune and it's about 80kg. You don't need to solve for a different mass on Neptune. Mass doesn't change. Only the weight changes.
Okay, I see what you're saying. I made a note in my notebook regarding universality of mass too, haha. I can't believe I skipped past that.

So if I plug the value for Mplanet of Neptune, Mperson, r^2 of Neptune, and G, then I will get around 897.06 N. I just convert this into pounds right?
 
  • #4
Dick
Science Advisor
Homework Helper
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Okay, I see what you're saying. I made a note in my notebook regarding universality of mass too, haha. I can't believe I skipped past that.

So if I plug the value for Mplanet of Neptune, Mperson, r^2 of Neptune, and G, then I will get around 897.06 N. I just convert this into pounds right?
Sure, but as you said, just using ratios is the easy way to do it.
 
  • #5
31
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Sure, but as you said, just using ratios is the easy way to do it.
Thank you!
 

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