Weight above the surface of the Earth

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Homework Help Overview

The problem involves calculating the weight of an astronaut at a specific altitude above the Earth's surface, given their weight on the surface. The subject area includes gravitational force and the application of Newton's law of universal gravitation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the correct equations to use, with some questioning the setup of the gravitational force formula. There are attempts to calculate the weight at altitude, with discussions on the necessary adjustments for distance from the Earth's center.

Discussion Status

Participants are actively engaging with the problem, providing calculations and corrections. There is a recognition of the need to adjust the distance used in the gravitational formula, and some guidance has been offered regarding the conversion of units and the implications of the inverse square law.

Contextual Notes

There is a mention of the astronaut's weight on the surface and the need to consider the radius of the Earth in calculations. The discussion includes clarifications on the proper interpretation of distance in the gravitational formula.

Sligh
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Homework Statement



An astronaut weighs 800 N on the surface of the Earth. What is the weight of the astronaut 6.37*10^6 km above the surface of the earth?

Homework Equations



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

W=ma

The Attempt at a Solution



I am sure that I need the previous equations I listed, although I am not quite sure how to actually solve this problem. A solution to this type of problem would be most appreciated. Thanks!
 
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you have your second equation wrong, it should read:
F = ma

G is constant, and you know d. So all you need are the two masses. Earth and astronaut.
 
Thanks for the assistance. Here's what I figured ->

F = 6.67E-11*(81.632kg)(5.9742E24kg) / (6.37E6)^2

F=801.624N

W=81.8014kg
Is this alright? Thanks!
 
Last edited:
Sligh said:
Thanks for the assistance. Here's what I figured ->

F = 6.67E-11*(81.632kg)(5.9742E24kg) / (6.37E6)^2

This would not give the weight at the point they are asking about. The d in the denominator of your equation is the distance to the center of the earth. So what would d be when the astronaut is at the height given?
 
Ah, you are correct. Then I must add the radius of the Earth plus the distance at which the astronaut is above the Earth, yes?

Here's my new results:

F = 6.67E-11*(81.632)(5.9742E24) / (6.37E6 + 6378.1)^2

F = 800.0514292 N

W = 81.63790094 kg

Is that it?

Thanks.
 
Sligh said:
Ah, you are correct. Then I must add the radius of the Earth plus the distance at which the astronaut is above the Earth, yes?

Here's my new results:

F = 6.67E-11*(81.632)(5.9742E24) / (6.37E6 + 6378.1)^2

Not quite; the Earth's radius is 6378.1 km, but you need to convert that to meters.
 
Got it. Thanks a lot!
 
You're welcome!

Notice in this problem that what they have done is set the altitude essentially equal to the Earth radius, so really the astronaut is doubling his distance from the center of the earth. In an inverse square law like gravity, if you double the separation, the force magnitude goes to one-fourth of its original value.
 

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