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yhzhang · Sep 18, 2012, 09:08 AM

Can someone help me with the question, I really don't know how to do it. Thanks in advance.
The Problem as follow:
Imagine that it is possible to take a 6 kg mass and raise it straight up off the Earth's equator on a huge tower that stretches beyond the Earth's atmosphere into space. If the top of the tower is 160.0 km directly above a point on the Earth's equator, what would the mass weight at the top of the tower?
For comparison, on the surface of the Earth, the mass weighs 59.2 Newtons. DATA: Equatorial radius of the Earth 6.378E6 meters; mass of the Earth 5.98E24 kg; Gravitational Constant 6.673E-11 Nm^2/kg^2

ebaines · Sep 18, 2012, 09:28 AM

The weight of the object is equal to the gravitational attraction of it to the earth. The formula for that is:

$ F_g = \frac {GMm}{R^2} $

You have been given the data for G,M (mass of the earth), and m (mass of the object). The value for R (distance from the object to the center if the earth) can be calculated by adding the earth's radius to the height above the earth's surface. Post back with what you get for an answer. Hint - the answer will be just a few percent less than the object's weight at the earth's surface.

yhzhang · Sep 18, 2012, 09:51 AM

I got the answer 0.0000063259, and i don't think this is the right answer. Also, I am not sure about the unit.

Originally Posted by ebaines

The weight of the object is equal to the gravitational attraction of it to the earth. The formula for that is:

$ F_g = \frac {GMm}{R^2} $

You have been gived the data for G,M (mass of the earth), and m (mass of the object). can The value for R (distance from the object to the center if the earth) can be calculated by adding the earth's radius to the height above the earth's surface. Post back with what you get for an answer. Hint - the answer will be just a few percent less than the object's weight at the earth's surface.

ebaines · Sep 18, 2012, 10:00 AM

Originally Posted by yhzhang

I got the answer 0.0000063259, and i don't think this is the right answer. Also, I am not sure about the unit.

Check your math - you haven't calculated GMm/R^2 correctly. As for units - if you write the equation out you should see that you end up with newtons:

$ F = \frac {GMm}{R^2} = \frac {(Nm^2/kg^2)(Kg)(Kg)}{m^2} = N $

yhzhang · Sep 18, 2012, 02:03 PM

Originally Posted by ebaines

Check your math - you haven't calculated GMm/R^2 correctly. As for units - if you write the equation out you should see that you end up with newtons:

$ F = \frac {GMm}{R^2} = \frac {(Nm^2/kg^2)(Kg)(Kg)}{m^2} = N $

Thank you, I got the right answer, which is 56.01 N