#### Pyrrhus

Homework Helper
If you put a ladder throught the Earth going throught its center, and if you go down it with your head pointing to the Earth's surface and feet to the centroid, passing throught Earth's centroid and heading to the other surface, you will go out with your head pointing to the surface and feet to the Earth's centroid. Why?

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#### Doc Al

Mentor
Because unless you are comfortable climbing a ladder upside down (with respect to the gravitation field) you will be motivated to turn around.

#### vaishakh

Lets discuss what is hapenning at the centroid. Coz that is where the question really matters? Please dor eply.

#### Doc Al

Mentor
The gravitational field at any radius always points towards the center of the earth. The effective acceleration due to gravity varies (linearly) from 0 at the center to the usual value of g at the earth's surface.

#### HallsofIvy

Well, it say "heading to the surface" so obviously you turn your head to the surface when you pass the center of the earth!

#### Doc Al

Mentor
:rofl: I think you've cracked this riddle!

#### Pyrrhus

Homework Helper
hahahah, Thanks for the replies. I was just wondering if it was something special or whatever.

#### cotton

wouldn't the gravitational pull increase as you neared the center of the earth? according to the law of universal gravitation, wouldn't r approach 0 and thus when you are at the center the gravitational force would be infinite?

#### Doc Al

Mentor
cotton said:
wouldn't the gravitational pull increase as you neared the center of the earth? according to the law of universal gravitation, wouldn't r approach 0 and thus when you are at the center the gravitational force would be infinite?
No. The gravitational field at the center is zero.

The gravitational pull at a distance r from the earth's center is only due to the mass contained within the sphere of radius r. (The outer shell contribution cancels.) While the the gravitational pull does increase according to M/r^2, M is proportional to r^3: the net effect is that the effective field strength is proportional to r. It's zero at the center (where r = 0) and the usual g at the surface.

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