Gravity/potential energy

Homework Statement

A sensitive gravimeter is carried to the top of Chicago’s Willis (formerly Sears) Tower, where its reading for the acceleration of gravity is 1.36 mm/s^2 lower than at street level.

Find the building's height, h=?

The Attempt at a Solution

I started this problem with this:

g - g' = 0.00136 m/s^2

now find the change in potential energy of gravity:

using this we should be able to simplify but i am not getting too far from here, it is a bit messy and i am not getting too much to cancel out here...why?

can anyone help me out?

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Doc Al
Mentor
I started this problem with this:

g - g' = 0.00136 m/s^2
OK.

now find the change in potential energy of gravity:
You're finding the change in g, not gravitational PE.

OK, except that it should be only GM in the numerator, not GMm.

using this we should be able to simplify but i am not getting too far from here, it is a bit messy and i am not getting too much to cancel out here...why?
Assume that h is small enough compared to the earth's radius that you can use a binomial expansion approximation.

Hint: (R + h)^2 = R^2(1 + h/R)^2; now take advantage of the fact that h/R << 1.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Look at $\displaystyle \frac{1}{R^2}-\frac{1}{(R+h)^2}=\frac{(R+h)^2-R^2}{R^2(R+h)^2}$

The numerator of the expression on the right side of the equation is the difference of squares.

Factor that.

R > 6300 km. I doubt that the building is more than 1 km tall (≈5/8 mile).

why is m not included in the formula?

going off the rest you said:

GM/(Re)^2 - GM/(Re + h)^2

Gm[(Re + h)^2 - (Re)^2] /(Re)^2 (Re + h)^2

using this: (R + h)^2 = R^2(1 + h/R)^2

i reduce down to :

Gm [ h^2/Re^2 + 2h/Re] / Re^2

does this sound right? still seems like a lot going on

Look at $\displaystyle \frac{1}{R^2}-\frac{1}{(R+h)^2}=\frac{(R+h)^2-R^2}{R^2(R+h)^2}$

The numerator of the expression on the right side of the equation is the difference of squares.

Factor that.

R > 6300 km. I doubt that the building is more than 1 km tall (≈5/8 mile).
Did you cancel out GM?

Doc Al
Mentor
why is m not included in the formula?
Because you want the acceleration, not the force.

going off the rest you said:

GM/(Re)^2 - GM/(Re + h)^2

Gm[(Re + h)^2 - (Re)^2] /(Re)^2 (Re + h)^2

using this: (R + h)^2 = R^2(1 + h/R)^2

i reduce down to :

Gm [ h^2/Re^2 + 2h/Re] / Re^2
Looks good. Now get rid of higher order terms, like (h/Re)2. They are too small to worry about. (And that m should be M, the mass of the earth.)

does this sound right? still seems like a lot going on
You're on the right track. Keep going.

right. now i get :

2GMh/Re^3

and get with an answer of 441.4 meters

which turns out to be the right answer, sweet. Thanks everyone!