Weight above the surface of the Earth

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SUMMARY

An astronaut weighing 800 N on Earth's surface experiences a weight of approximately 800.05 N when positioned 6.37 million meters above the surface. The calculations utilize Newton's law of universal gravitation, specifically the formula F = G (m1)(m2)/(d)^2, where G is the gravitational constant. The correct distance (d) used in the calculations must include the Earth's radius, leading to a final weight calculation of W = 81.6379 kg. Understanding the inverse square law of gravity is crucial, as doubling the distance from the Earth's center reduces the gravitational force to one-fourth.

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  • Newton's law of universal gravitation
  • Understanding of gravitational force calculations
  • Knowledge of unit conversions (kilometers to meters)
  • Familiarity with the concept of inverse square law
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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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