Tunnel drilled through the Earth and mechanical work

Click For Summary
SUMMARY

The discussion centers on calculating the mechanical work done by gravitational force when moving a 1.0 kg mass through a tunnel drilled straight through the Earth to a midpoint located at 0.8 times the Earth's radius (RE). The user initially calculated the work as W = 1.1258 × 107 J, while the reference book states the answer should be W = 5.1 × 107 J. The discrepancy arises from the incorrect application of the gravitational force equation F = -GmMEd/RE3, which does not account for the varying mass of the Earth within the radius considered.

PREREQUISITES
  • Understanding of gravitational force equations, specifically F = -GmMEd/RE3.
  • Knowledge of gravitational constant G = 6.673 × 10-11 m3 kg-1 s-2.
  • Familiarity with Earth's mass ME = 5.98 × 1024 kg and radius RE = 6.38 × 106 m.
  • Basic principles of mechanical work and energy conservation in physics.
NEXT STEPS
  • Review the derivation of gravitational force equations in spherical coordinates.
  • Study the concept of gravitational potential energy and its relation to work done against gravity.
  • Learn about the implications of varying mass distribution within celestial bodies.
  • Explore advanced topics in gravitational physics, such as the Shell Theorem.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and gravitational forces, as well as educators looking for examples of gravitational work calculations.

Kolossi
Messages
1
Reaction score
0

Homework Statement


Suppose that a straight tunnel is drilled through the Earth as indicated in figure. Assume that the floor of the tunnel is frictionless and that air resistance can be neglected. If the mid-point of the tunnel is 0.8[tex]R_{E}[/tex] from the centre of the Earth, where the [tex]R_{E}[/tex] is the radius of the Earth, determine the mechanical work done by the field force in moving a 1.0 kg mass from the entrance of the tunnel to the mid-point.

Homework Equations


[tex]F=-GmM_{E}d/R_{E}^3[/tex], where the d is distance from the centre of the Earth

The Attempt at a Solution


http://img35.imageshack.us/img35/323/soltry.jpg

[tex]G\ =\ 6.673(10)\ \times\ 10^{-11}\ m^{3} kg^{-1} s^{-2}[/tex]
[tex]M_{E}\ =\ 5.98\ \times\ 10^{24}\ kg[/tex]
[tex]R_{E}\ =\ 6.38\ \times\ 10^{6}\ m[/tex]
[tex]m\ =\ 1.0 kg[/tex]

I get [tex]W\ =\ 1.1258\ \times\ 10^{7}\ J[/tex]

According to book [tex]W[/tex] should be [tex]5.1\ \times\ 10^{7}\ J[/tex]

What did I do wrong? Or is the answer in the book wrong?

Thank you for in advance!
 
Last edited by a moderator:
Physics news on Phys.org
Kolossi said:

Homework Statement


Suppose that a straight tunnel is drilled through the Earth as indicated in figure. Assume that the floor of the tunnel is frictionless and that air resistance can be neglected. If the mid-point of the tunnel is 0.8[tex]R_{E}[/tex] from the centre of the Earth, where the [tex]R_{E}[/tex] is the radius of the Earth, determine the mechanical work done by the field force in moving a 1.0 kg mass from the entrance of the tunnel to the mid-point.



Homework Equations


[tex]F=-GmM_{E}d/R_{E}^3[/tex], where the d is distance from the centre of the Earth


The Attempt at a Solution


http://img35.imageshack.us/img35/323/soltry.jpg

[tex]G\ =\ 6.673(10)\ \times\ 10^{-11}\ m^{3} kg^{-1} s^{-2}[/tex]
[tex]M_{E}\ =\ 5.98\ \times\ 10^{24}\ kg[/tex]
[tex]R_{E}\ =\ 6.38\ \times\ 10^{6}\ m[/tex]
[tex]m\ =\ 1.0 kg[/tex]

I get [tex]W\ =\ 1.1258\ \times\ 10^{7}\ J[/tex]

According to book [tex]W[/tex] should be [tex]5.1\ \times\ 10^{7}\ J[/tex]

What did I do wrong? Or is the answer in the book wrong?

Thank you for in advance!

Not sure your first equation for F is correct. You are applying a linear ratio to the radius distance, but the mass of the part of the Earth inside your radius varies with the cube of the radius, doesn't it? Maybe revisit how you set up that first equation and see if that fixes the solution.

Welcome to the PF, BTW.
 
Last edited by a moderator:

Similar threads

Replies
15
Views
2K
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 15 ·
Replies
15
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 6 ·
Replies
6
Views
6K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 24 ·
Replies
24
Views
3K
Replies
37
Views
4K