Gravitational Force Problem Solution

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SUMMARY

The gravitational force problem discusses the weight loss of a student weighing 750 N at the Earth's surface when descending 11.8 km into a nonrotating uniform sphere. The key equation used is F = G Mm/R^2, where G is the gravitational constant (6.67 * 10^-11 Nm^2/kg^2), M is the Earth's mass (5.98 * 10^24 kg), and R is the radius adjusted for depth. The solution reveals that as one descends, the effective mass of Earth below decreases, leading to a reduction in gravitational force, resulting in a weight loss of approximately 2.77 N at the bottom of the shaft.

PREREQUISITES
  • Understanding of gravitational force equations, specifically F = G Mm/R^2
  • Knowledge of Earth's mass and radius
  • Familiarity with concepts of density and uniform spheres
  • Basic algebra for manipulating equations and calculating weight
NEXT STEPS
  • Study gravitational force calculations in varying depths using "Gravitational Field Strength" concepts
  • Learn about the implications of density variations in planetary bodies
  • Explore the effects of Earth's rotation on gravitational force
  • Investigate real-world applications of gravitational experiments in geophysics
USEFUL FOR

Physics students, educators, and anyone interested in gravitational theory and its applications in real-world scenarios, particularly in geophysics and planetary science.

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Homework Statement



Suppose Earth were a nonrotating uniform sphere. As a reward for earning the highest lab grade, your physics professor chooses your laboratory team to participate in a gravitational experiment at a deep mine on the equator. This mine has an elevator shaft going 11.8 km into Earth. What would be the loss in weight at the bottom of this deep shaft for a student who weighs 750 N at the surface of the Earth?

Homework Equations



F = G Mm/R^2

The Attempt at a Solution



Well from the beginning, the problem confuses me because wouldn't your weight increase if you go deeper into the earth? Since that means R would be decreased?

Anyways, plugging in the values for M (ignoring m since it's so small in comparison) as 5.98 * 10^24 kg, R as (6380 kg - 11.8 km), and G as 6.67 * 10^-11 Nm^2/kg^2, I get a value of 9.8354... Then I divide that by gravity on the surface (9.8 m/s^2) to find how much the weight will increase. However, I cannot get the right answer!

Submission # Try Submitted Answer
1 Submission not graded. Use more digits. 2.8 N
2 Incorrect. (Try 1) 2.79 N
3 Incorrect. (Try 2) -2.79 N
4 Submission not graded. Use more digits. 3 N
5 Incorrect. (Try 3) 3.00 N
6 Submission not graded. Use more digits. 3. 5N
7 Submission not graded. Use more digits. 3. 5 N
8 Incorrect. (Try 4) 2.77 N
9 Incorrect. (Try 5) -2.77 N
10 Incorrect. (Try 6) 2.78 N
11 Incorrect. (Try 7) -2.78 N

Thank you for any help!
 
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In the real world, you are correct. Your weight increases as you go down, until you reach the edge of the core. This is because 2 things are happening:

1. You are getting closer to the center of Earth.
2. Only the mass below you contributes to your acceleration.

So by going down, your Earth effectively loses mass, causing gravity to weaken. But you get closer to the center of mass, causing your gravity to strengthen. Since Earth's core is denser than it surface, in the real world, your decreasing r wins, and you get heavier. But in a uniform sphere, the decreasing effective mass wins, and you get lighter.

So in this problem, you only consider the mass of the Earth below you. Compute the density of the Earth using its radius and mass, then using this density, compute the mass of an Earth that is 11.8 km smaller in radius. Use that mass and your new r to compute your new acceleration due to gravity. Then you can compute your weight loss.
 
Thank you! I got the answer now.
 

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