Weight Away from Earth's Surface [CONCEPT-NO MATH]

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SUMMARY

This discussion focuses on calculating the weight of a spacecraft at varying distances from Earth's surface, specifically at 6.38x10^3 km and 1.28x10^4 km. The key formula used is F = mAmB/d^2, where d represents the distance from the center of the Earth, not the distance to the surface. The confusion arises from the interpretation of distances as rE + rE and rE + 2rE, where rE is the Earth's radius. The conclusion emphasizes that the distance for gravitational force calculations must account for the total distance from the Earth's center.

PREREQUISITES
  • Understanding of gravitational force equations, specifically F = mAmB/d^2
  • Familiarity with Earth's radius (rE) and its significance in gravitational calculations
  • Basic knowledge of spacecraft weight calculations in relation to distance from Earth
  • Concept of gravitational force variation with distance from a massive body
NEXT STEPS
  • Study the implications of gravitational force variations with distance using Newton's Law of Universal Gravitation
  • Explore the concept of gravitational potential energy and its relationship to distance from Earth's center
  • Learn about the effects of altitude on spacecraft weight and performance
  • Investigate the calculations involved in orbital mechanics and spacecraft trajectory planning
USEFUL FOR

Aerospace engineers, physics students, and anyone interested in understanding gravitational effects on spacecraft at varying distances from Earth.

SigFig
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Homework Statement



When calculating weights away from Earth's surface, for example:

At 6.38x10^3 km away from Earth's surface a spacecraft s weight is d=rE+rE or 2rE
Then F=(1/4)(SpaceCraftWeight)

What I don't understand is why the distance is rE+rE?

Or when 1.28x10^4 km away
d=rE+2rE or 3rE ?

I'm also assuming since F=mAmB/d^2
That's why F= (1/(x^2)(W) ?
 
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SigFig said:

Homework Statement



When calculating weights away from Earth's surface, for example:

At 6.38x10^3 km away from Earth's surface a spacecraft s weight is d=rE+rE or 2rE
Then F=(1/4)(SpaceCraftWeight)

What I don't understand is why the distance is rE+rE?

Or when 1.28x10^4 km away
d=rE+2rE or 3rE ?

I'm also assuming since F=mAmB/d^2
That's why F= (1/(x^2)(W) ?

Because the d in F=mAmB/d^2 is the distance from the center of the earth. Not the distance to the Earth's surface.
 

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