How High Above Earth is Gravity 10% of Sea Level?

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SUMMARY

The discussion focuses on determining the height above Earth's surface where gravitational acceleration is 10% of that at sea level. The relevant equation is derived from Newton's law of universal gravitation, expressed as F = Gm1m2/r². The solution involves manipulating the equation to find the height (x) where the gravitational force per unit mass equals 0.1 times the gravitational acceleration at sea level, leading to the equation GM/(r+x)² = 0.1 * GM/r².

PREREQUISITES
  • Understanding of Newton's law of universal gravitation
  • Familiarity with gravitational acceleration concepts
  • Basic algebra for manipulating equations
  • Knowledge of Earth's radius (approximately 6,371 kilometers)
NEXT STEPS
  • Calculate the height above Earth's surface using the equation GM/(r+x)² = 0.1 * GM/r²
  • Explore gravitational force variations with altitude
  • Study the implications of gravitational changes on satellite orbits
  • Learn about gravitational potential energy and its relationship with height
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Students in physics, educators teaching gravitational concepts, and anyone interested in the effects of altitude on gravitational forces.

ryryguy
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Homework Statement


At what height above the Earth's surface is the acceleration due to gravity 10% of that at sea level?

Homework Equations


F= Gm1m2/r(squared)


The Attempt at a Solution


I think some how one of the masses is moved over so that F/m=a and I think .10 is multiplied times something like (r squared) or (r + x)squared.
 
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As you said, F/m=a. When at the surface of the Earth at sea level:

[tex]\frac{F}{m}=\frac{GMm}{m(r+x)^2}=\frac{GM}{x^2}[/tex]

where x= radius of Earth, and since at the sea level r would be zero.

Now what is 10% of this? After you find this, can you set up an equation and solve for the distance from Earth? If you need more help feel free to ask. Good Luck!
 

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