# Homework Help: Change in weight

1. Oct 29, 2015

I attached picture.

I took the F=GMm/r^2 equation and did it for both Earth level, (radius of earth) with 650/9.8 as m

and for F at the top of the building, with r being the radius of the earth plus a mile (1.609E3 m).

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2. Oct 29, 2015

### SteamKing

Staff Emeritus
For this problem, W = F = GMm / r2,

where
M - mass of the earth,
G - Universal Gravitational constant

By changing elevation, does G, M or m change?

3. Oct 29, 2015

Sorry if I didn't include enough on this question. I thought the attached photo was enough but I guess not.

I assumed that the only change would be the r^2 in the equation.

So I put it through the eq. F = GMm / r2
but just increased the radius. Now I know the weight would change as the rider moved, but it is merely asking for the difference between the top and bottom (earth level) of the building so I would assume (hopefully correctly) that 'm' is static and only changes in reference to distance from the Ecom

4. Oct 29, 2015

Etop=6.379609E6

I got for bottom 649.651N and 649.324N... but that isn't right...

5. Oct 29, 2015

### haruspex

6. Oct 29, 2015

### SteamKing

Staff Emeritus
Your weight is 650 N on the ground, period. By how much will this number change 1609 meters up in the air?

You don't have to fool with G, M, or m, because all these number cancel out.

7. Oct 29, 2015

### haruspex

You have to be careful when the calculation involves finding a small difference between large numbers. Any little rounding errors can lead to major inaccuracy in the answer.
A safer way to proceed is to think in terms of fractional change. A change of 1 mile in roughly 4000 miles (no need to convert to km) is what fraction? Since the distance gets squared, how big a fraction should the change in weight be?

8. Oct 29, 2015

Problem in text format: In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constructed. Ignoring Earth's rotation, find the change in your weight if you were to ride an elevator from the street level, where you weigh 650 N, to the top of the building.

@SteamKing > I assume it means weight in reference to gravitational pull of the earth versus overall mass? I am not sure. Example answers from chegg and yahoo answers correlate with the answer I am getting. ..

@haruspex > Fearth = ((6.67E-11)(650/9.8)(5.97E24))/(6.378E6)^2

Ftop = ((6.67E-11)(650/9.8)(5.97E24))/6.379609E6)^2

9. Oct 29, 2015

### haruspex

Yes, I figured out from your post #4 what you had done, but didn't see that until my post 5. Can you answer my post 7?

10. Oct 29, 2015

Ah, a change of 1 in 4000 is 0.00025?

11. Oct 29, 2015

### haruspex

Right, so what fractional change should that lead to in weight?

12. Oct 29, 2015

650N >...0.1625N change...but its squared...so would change double to 0.325?

13. Oct 29, 2015

### haruspex

Yes.

14. Oct 29, 2015

So...what radius should I use...4000 is ok but with only 2% available lee-way I should use the most accurate possible. I do not like how it is not given in the problem...

so google says 3959miles...so using fractional change =0.32836...etc.

15. Oct 29, 2015

which it tells me is wrong..... :/

16. Oct 29, 2015

### haruspex

OK. 4000 had about a 1% error. The doubling of the fractional change does not alter the percentage error in the fractional change, so it should be within the allowed margin.

17. Oct 29, 2015

### haruspex

Does it need a signed answer?

18. Oct 29, 2015

ok...so I will try 4000, with 0.325??

19. Oct 29, 2015

...not that I know of??

These online assignments are very disheartening....

20. Oct 29, 2015

### haruspex

If it didn't like .328 I don't believe it will be happy with .325.

21. Oct 29, 2015

I agree with you there.... but I have no clue why..

22. Oct 29, 2015

### haruspex

Sure, but that was not asking about a change in force. Try -.328.

23. Oct 29, 2015

It took it. So is that typically universal when it comes to gravitational bodies and moving away from them? I mean it makes sense..I move away, I weigh less, so the change in force is always negative as I move away.

I am sure you could complicate that if the body being acted on was placed between two gravitational bodies...then I am assuming +and - would be in relation to the reference point, or what G you were using...

Just trying to make sure I understand the concepts..

24. Oct 29, 2015