Weight at different elevations in a tall building

[elissadi]

I have done most of the work on the problem written below, but the more I look at it, the more I am not sure whether I should leave the pounds as they are instead of converting them to kilograms. If I leave it as 200 lbs, are the number plug-ins going to be wrong?
I am using the formula
g = 6.67x10^-11 x 6 x 10^24 / (2 x 6.40 x 10^3)^2 = 2.44 m/s^2 then W=mg which I have calculated at 20 x 2.44 = 48.8 N
A building is 6400 km high. On the bottom floor a person weighs 200 pounds when stepping on a spring scale. How much would the person weigh on the same scale if she were standing on the top floor?
I know that 6400 km is also the radius of the Earth so that the top floor is 2R from the Earth's center.
thanks!

 

[Doc Al]

If you are going to "plug in" numbers in a formula like W = mg, then you surely better use mass and not weight!

But for this problem you should be able to get the answer using ratios without plugging any numbers in. By what factor does the gravitational force change when the object goes from 1R to 2R?

 

[elissadi]

so, I should divide 200/9.8m/s^2 to get the mass which would be 20 kgs. Or should I convert 200 lbs to kilograms first?

 

[AKG]

Let F represent the weight of the person on the floor
Let F' represent his weight on the roof
Let m represent his mass
Let M represent the Earth's mass
Let R represent the Earth's radius
Let G represent the gravitational constant

F = GmM/R²
F' = GmM/(2R)²

you have 2 equations and 2 unknowns (F' and m). Don't touch any numbers. Do the algebraic manipulations first, then plug in F = 200 lbs (you will never need to touch the numbers like G, M, or R -- although you did need to know that 6400 km = R -- nor will you need to convert any units). So what answer are you getting?

 

[BobG]

so, I should divide 200/9.8m/s^2 to get the mass which would be 20 kgs. Or should I convert 200 lbs to kilograms first?

You should convert the 200 lb to kilograms first.

You could divide 200 lb by 32 f/s to convert to mass in slugs, but you'd still be in the wrong units. You'd then have to convert slugs to kilograms. Pounds, feet, and slugs are measurements in the 'English' system and kilograms, meters, and Newtons are in the 'metric' system.

What the others are trying to tell you is that, in this particular problem, there's a better way to solve the problem than just plugging numbers into a memorized formula.

 

[Severian596]

Let F represent the weight of the person on the floor
Let F' represent his weight on the roof
Let m represent his mass
Let M represent the Earth's mass
Let R represent the Earth's radius
Let G represent the gravitational constant

F = GmM/R²
F' = GmM/(2R)²

you have 2 equations and 2 unknowns (F' and m). Don't touch any numbers. Do the algebraic manipulations first, then plug in F = 200 lbs (you will never need to touch the numbers like G, M, or R -- although you did need to know that 6400 km = R -- nor will you need to convert any units). So what answer are you getting?

I like this advice! Nice approach. You know that the force of gravity is not directly proportional to the distance between the objects. Rather it is...some other proportion

 

[elissadi]

Thank you all! I think all this advice will help greatly! I will work on this when I get back from class :)