Weight Change: Earth vs Building Top | F=GMm/r^2

[SadPanda6022]

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).

Please help?

 

[SteamKing]

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).

Please help?

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?

 

[SadPanda6022]

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

 

[SadPanda6022]

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...

 

[haruspex]

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).

Please help?

Your answer is just a bit inaccurate, but I'm not able to follow your working from this bare description. Please post the details.

 

[SteamKing]

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...

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.

 

[haruspex]

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...

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?

 

[SadPanda6022]

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

 

[haruspex]

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

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]

Ah, a change of 1 in 4000 is 0.00025?

 

[haruspex]

Ah, a change of 1 in 4000 is 0.00025?

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

 

[SadPanda6022]

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

 

[haruspex]

its squared...so would change double to 0.325?

Yes.

 

[SadPanda6022]

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.

 

[SadPanda6022]

which it tells me is wrong... :/

 

[haruspex]

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.

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.

 

[haruspex]

which it tells me is wrong... :/

Does it need a signed answer?

 

[SadPanda6022]

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

 

[SadPanda6022]

...not that I know of??

These online assignments are very disheartening...

 

[haruspex]

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

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

 

[SadPanda6022]

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

The sample problem linked yields a positive answer.

 

[haruspex]

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

The sample problem linked yields a positive answer. View attachment 91048

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

 

[SadPanda6022]

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..

 

[SadPanda6022]

THANKS BY THE WAY!

 

[haruspex]

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..

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]

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!