How Does Distance from Earth Affect Spacecraft Weight?

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Homework Help Overview

The discussion revolves around the effect of distance from Earth on the weight of a spacecraft, specifically examining a spacecraft with a weight of 7200 N at varying distances from the Earth's surface, including 6400 km and 12800 km. The problem involves gravitational concepts and calculations based on the universal law of gravitation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the weight of the spacecraft using gravitational formulas, questioning whether their approach is correct. Some participants suggest simplifying the problem by considering the factors by which the distance increases and how this affects gravitational force.

Discussion Status

Participants are exploring different interpretations of the problem, with some suggesting a simpler method based on the inverse square law of gravitation. There is no explicit consensus, but guidance has been offered regarding the relationship between distance and gravitational force.

Contextual Notes

Participants note the significance of the given distances and the inverse square law, indicating a focus on understanding the implications of distance on gravitational force without resolving the calculations completely.

Coco12
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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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