- #1
Weight Change: Earth vs Building Top | F=GMm/r^2
Click For Summary
Homework Help Overview
The discussion revolves around the gravitational force experienced by an individual at different elevations, specifically comparing weight at Earth's surface to weight at the top of a hypothetical mile-high building. The relevant equation is F=GMm/r^2, where G is the gravitational constant, M is the mass of the Earth, m is the mass of the individual, and r is the distance from the center of the Earth.
Discussion Character
- Exploratory, Assumption checking, Mathematical reasoning
Approaches and Questions Raised
- Participants explore the implications of changing the radius in the gravitational force equation and question whether other variables (G, M, m) remain constant. There are discussions about the accuracy of calculations and the significance of small changes in weight due to elevation.
Discussion Status
Participants are actively engaging with the problem, sharing calculations and questioning the assumptions made regarding the variables in the gravitational force equation. Some have provided guidance on how to approach the problem, particularly regarding the importance of considering fractional changes and potential rounding errors in calculations.
Contextual Notes
There is mention of constraints related to homework rules and the need for precise values, as well as the challenge of interpreting the problem's requirements regarding the change in weight.
- #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
- 12,828
- 1,673
SadPanda6022 said:
For this problem, W = F = GMm / r2,
where
M - mass of the earth,
m - your mass,
G - Universal Gravitational constant
By changing elevation, does G, M or m change?
- #3
SadPanda6022
- 33
- 0
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
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
SadPanda6022
- 33
- 0
so, I had Emass=5.9736E24
Eradius=6.378E6
add on a mile
Etop=6.379609E6
I got for bottom 649.651N and 649.324N... but that isn't right...
Eradius=6.378E6
add on a mile
Etop=6.379609E6
I got for bottom 649.651N and 649.324N... but that isn't right...
- #5
- 42,884
- 10,519
Your answer is just a bit inaccurate, but I'm not able to follow your working from this bare description. Please post the details.SadPanda6022 said:
- #6
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
- 12,828
- 1,673
Your weight is 650 N on the ground, period. By how much will this number change 1609 meters up in the air?SadPanda6022 said:
You don't have to fool with G, M, or m, because all these number cancel out.
- #7
- 42,884
- 10,519
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.SadPanda6022 said:
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
SadPanda6022
- 33
- 0
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
@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
- 42,884
- 10,519
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?SadPanda6022 said:
- #10
SadPanda6022
- 33
- 0
Ah, a change of 1 in 4000 is 0.00025?
- #11
- 42,884
- 10,519
Right, so what fractional change should that lead to in weight?SadPanda6022 said:
- #12
SadPanda6022
- 33
- 0
650N >...0.1625N change...but its squared...so would change double to 0.325?
- #13
- 42,884
- 10,519
Yes.SadPanda6022 said:
- #14
SadPanda6022
- 33
- 0
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.
so google says 3959miles...so using fractional change =0.32836...etc.
- #15
SadPanda6022
- 33
- 0
which it tells me is wrong... :/
- #16
- 42,884
- 10,519
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.SadPanda6022 said:
- #17
- 42,884
- 10,519
Does it need a signed answer?SadPanda6022 said:
- #18
SadPanda6022
- 33
- 0
ok...so I will try 4000, with 0.325??
- #19
SadPanda6022
- 33
- 0
...not that I know of??
These online assignments are very disheartening...
These online assignments are very disheartening...
- #20
- 42,884
- 10,519
If it didn't like .328 I don't believe it will be happy with .325.SadPanda6022 said:
- #21
SadPanda6022
- 33
- 0
I agree with you there... but I have no clue why..
The sample problem linked yields a positive answer.
The sample problem linked yields a positive answer.
- #22
- 42,884
- 10,519
Sure, but that was not asking about a change in force. Try -.328.SadPanda6022 said:
- #23
SadPanda6022
- 33
- 0
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..
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
SadPanda6022
- 33
- 0
THANKS BY THE WAY!
- #25
- 42,884
- 10,519
Yes, it would need to make the reference direction clearer in that case. Of course, you could argue that you were given the reference direction in the statement that the ground level weight is 650N, not -650N. The change would then be clearly negative.SadPanda6022 said:
- #26
SadPanda6022
- 33
- 0
I was going to come back and give you praise somehow but...I have no idea how forums work as this is the first one I have ever frequented. So I gave you a measly message like.
Thanks for the help!
Thanks for the help!