How Does Distance from Earth Affect Spacecraft Weight?

AI Thread Summary
The discussion focuses on calculating the weight of a spacecraft at varying distances from Earth's surface using gravitational formulas. The gravitational force is determined by the inverse square law, indicating that as distance increases, the force of gravity decreases proportionally to the square of the distance from Earth's center. For a spacecraft weighing 7200 N at Earth's surface, its weight would be reduced to one-fourth at 12800 km and one-ninth at 19200 km from the center. Participants confirm that using the radius of Earth plus the distance traveled is the correct approach for these calculations. The conversation emphasizes understanding the relationship between distance and gravitational force.
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Homework Statement



The radius of the Earth is 6400km. A 7200n spacecraft travels away from the earth. what would be the weight of the spacecraft at the following distances from the Earths surface: 6400km, 12800km

Homework Equations



Fg=mg
F=Gm1m2/r^2

The Attempt at a Solution


Just want to confirm I'm doing this right
So first I want to find the mass of the spacecraft so I use the fg=mg formula. I know that the mass of the Earth is 5.98*10^24kg.

Gravitational constant is 6.67*10^-11

I use the universal gravitation formula

The r will just be the radius of the Earth PLUS the distance from the Earth's surface. (Converted to m)

Then just plug into the equation. Am I doing this right?
 
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It is not wrong but too complicted. They give these nice numbers for a reason.
Look at the numbers. By what factor increases the distance? What does this tell about the change in force?
 
The factor increases by 2 (6400+6400) then 3 (6400+12800)

So u mean I would just multiply 7200 by 1/4 for the distance 12800 from th Earth's center
And 1/9 for the distance 19200 ?
 
Coco12 said:
The factor increases by 2 (6400+6400) then 3 (6400+12800)

So u mean I would just multiply 7200 by 1/4 for the distance 12800 from th Earth's center
And 1/9 for the distance 19200 ?

You have what is known as 'an inverse square law'

ie

Force due to gravity decreases with 1/r^2
 
Coco12 said:
The factor increases by 2 (6400+6400) then 3 (6400+12800)

So u mean I would just multiply 7200 by 1/4 for the distance 12800 from th Earth's center
And 1/9 for the distance 19200 ?

Yes. This will do it.
 
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