Calculating the Distance to Reach 7.5% Weight Reduction of a Uniform Sphere

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SUMMARY

The discussion focuses on calculating the depth required to achieve a 7.5% weight reduction when drilling into the Earth, modeled as a uniform sphere. The gravitational force equation, Fg=(GMm)/r², is utilized to derive the relationship between the surface weight and the weight at a depth. By setting the weight at depth to 92.5% of the surface weight, the problem simplifies to finding the radius at which this condition holds true. The solution involves solving for the radius 'r' at the specified weight reduction.

PREREQUISITES
  • Understanding of gravitational force equations, specifically Fg=(GMm)/r²
  • Knowledge of uniform sphere properties in physics
  • Basic algebra for solving equations
  • Familiarity with concepts of weight reduction and percentage calculations
NEXT STEPS
  • Study the implications of gravitational force inside a uniform sphere
  • Explore the derivation of gravitational equations for varying depths
  • Learn about the concept of weight reduction in gravitational fields
  • Investigate real-world applications of gravitational calculations in geophysics
USEFUL FOR

Students in physics, particularly those studying gravitational forces, as well as educators and anyone interested in the mathematical modeling of physical phenomena related to weight and depth in a uniform sphere context.

Charanjit
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1. Homework Statement :
It can be shown that for a uniform sphere the force of gravity at a point inside the sphere depends only on the mass closer to the center than that point. The net force of gravity due to points outside the radius of the point cancels.

Question: How far would you have to drill into the Earth, to reach a point where your weight is reduced by 7.5% ? Approximate the Earth as a uniform sphere.




2. Homework Equations :
Fg=(GMm)/r2




3. The Attempt at a Solution :

Well because the weight is reduced by 7.5 the inital weight has to be 0.925. This is a tough one. Thats why I need help. Get me started.
 
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You want the new weight to be 92.5% of the weight on the surface. Therefore you want0.925 \frac{GMm}{R^2} = \frac{GMm}{r^2}

What should R be equal to in this case?
 
So is R the radius of the earth? So GMm cancel and all we are left with is 0.925, R2 and r. Solve for r?
 
I got it solved. Thanks.
 

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