You are a tourist on space odyssey; at what altitude above the surface

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Homework Help Overview

The problem involves determining the altitude above Earth's surface at which a person's weight is half of their weight on the surface. The context is rooted in gravitational physics, specifically relating to the gravitational force equation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants express uncertainty about how to approach the problem, with one questioning whether the weight in question refers to actual or apparent weight. Others discuss the implications of gravitational constants and the variables involved in the gravitational force equation.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of weight and gravitational effects. Some guidance has been offered regarding the variables in the gravitational force equation, but no consensus or resolution has been reached.

Contextual Notes

Participants note the importance of distinguishing between actual and apparent weight, as well as the implications of being in free fall, which may affect the interpretation of the problem.

jasonbans
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Homework Statement


You are a tourist on space odyssey; at what altitude above the surface of Earth will your weight be one-half your weight on the surface? express your answer as a mutiple of Earth's radius rE


Homework Equations


Fg= GMm/r^2


The Attempt at a Solution


not sure how to do it
 
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jasonbans said:

Homework Statement


You are a tourist on space odyssey; at what altitude above the surface of Earth will your weight be one-half your weight on the surface? express your answer as a mutiple of Earth's radius rE


Homework Equations


Fg= GMm/r^2


The Attempt at a Solution


not sure how to do it

Do you mean your actual weight or your apparent weight? Remember the apparent weight of people on the space station is zero as they are in free fall, but their actual weight is around 90% of their weight on Earth - because there g ~ 8.9 rather than the 9.8 on the surface.
 


PeterO said:
Do you mean your actual weight or your apparent weight? Remember the apparent weight of people on the space station is zero as they are in free fall, but their actual weight is around 90% of their weight on Earth - because there g ~ 8.9 rather than the 9.8 on the surface.

actual weight
 


jasonbans said:
actual weight

Look at your formula and consider what change can be made to halve the force.

I don't think you can change G - is is after all the Universal Gravitational Constant.

M won't alter - mass of the Earth

You could halve m if you put the passenger on a diet, but I don't think that is feasible.

Doesn't leave much else - and remember R is the distance from the centre of the Earth, so we are already one Earth radius away when standing on the surface.
 

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