How much does a person weigh on Mount Everest?

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Homework Help Overview

The discussion revolves around calculating the weight of a person on Mount Everest, specifically addressing the effects of altitude on gravitational force. The original poster presents a scenario involving a person with a mass of 68 kg and seeks to determine their weight at an elevation of 8488 meters above sea level.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between mass and weight, questioning how altitude affects gravitational force. There are attempts to clarify the concept of the "old" radius in the gravitational formula and its relevance at different altitudes.

Discussion Status

The discussion is ongoing, with participants raising questions about the application of Newton's laws of gravity at different altitudes. Some have provided calculations for weight at sea level, while others are examining the implications of using the universal law of gravitation in this context. There is no explicit consensus on the approach to take, but various interpretations and clarifications are being explored.

Contextual Notes

Participants are navigating the complexities of gravitational calculations, particularly in relation to altitude changes. The original poster's problem involves specific values for mass and height, but there are questions about the assumptions made regarding gravitational strength at different elevations.

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Homework Statement



If a person has a mass of 68kg, how much does he weigh on the top of mount everest (8488km above sea level?
Given: m=68kg, d=8488m

Homework Equations



F = (G(m1)(m2)) / r^2

The Attempt at a Solution



I tried figuring out the new radius. Once i get the radius i enter all the data and rearrange the formula to get m2 by itself.
Fr^2/Gm1 = m2

My problem is getting the new radius.
 
Last edited:
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What's the "old" radius?
 
D H said:
What's the "old" radius?

Sorry, the old radius is the basic radius in the original formula. Its just r^2. But since I am dealing with a new altitude doesn't it change?
 
You didn't understand my question. I'll take another tack. How much does the person weigh at sea level? Why?
 
D H said:
You didn't understand my question. I'll take another tack. How much does the person weigh at sea level? Why?

He weighs 668N at sea level. Why? Well His mass multiplied by gravity gives me the weight.
 
What about Newton's law of gravity? Does that not apply at sea level?
 
D H said:
What about Newton's law of gravity? Does that not apply at sea level?

As long as an object is at or near the surface of the celestial object and we know the value of g the gravitational field strength at the surface of the celestial body, we can use F= mg to find the weight of the object
 
Newton's universal law of gravitation, F=GMm/r2, is not called universal just on a whim.

In other words, it applies at sea level as well as atop Mt. Everest.
 
D H said:
Newton's universal law of gravitation, F=GMm/r2, is not called universal just on a whim.

In other words, it applies at sea level as well as atop Mt. Everest.

But the answers are different. My answer book tells me that the weight on the surface of the Earth is 668 Newtons and 664 Newtons on mount Everest.
 
  • #10
Whoops, was i not supposed to read that xD?
 
Last edited:
  • #11
No, you were not supposed to see that.
 

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