# Math Help - Weight Elevation

1. ## Weight Elevation

If an object weighs m pounds at sea level, then its weight W (in pounds) at a height of h miles above seal level is given approximately by:

W(h) = m(4000)/(4000 + h)^2

If Ritter weighs 120 pounds at sea level, how much will he weigh on Pike's Peak, which is 14,110 feet above sea level?

2. Originally Posted by symmetry
If an object weighs m pounds at sea level, then its weight W (in pounds) at a height of h miles above seal level is given approximately by:

W(h) = m(4000)/(4000 + h)^2

If Ritter weighs 120 pounds at sea level, how much will he weigh on Pike's Peak, which is 14,110 feet above sea level?
The formula is wrong. I'm not going to bother looking up what the mass of the Earth and G are in English units and just do the problem as stated.

Well at sea level you know W = 120 lbs, and for h = 0, W = m.
There are 5280 ft in 1 mile, so:
$\frac{14110 \, ft}{1} \cdot \frac{1 \, mi}{5280 \, ft} = 2.67235 \, mi$

Thus
W(14110) = (120)(4000)/(4000 + 2.67235)^2 = 0.02996 lbs.

-Dan

3. ## Dan

Dan,

Thanks for your reply but why did you say the formula is wrong?

I got the formula from the textbook.

m = pounds

h = miles

W = weight (in pounds)

The formula is:

W(h) = m times the fraction (4000/4000 + h). The entire fraction is squared.

If this is not the correct formula, do you know what the correct formula looks like?

Thank!

4. Originally Posted by symmetry
Dan,

Thanks for your reply but why did you say the formula is wrong?

I got the formula from the textbook.

m = pounds

h = miles

W = weight (in pounds)

The formula is:

W(h) = m times the fraction (4000/4000 + h). The entire fraction is squared.

If this is not the correct formula, do you know what the correct formula looks like?

Thank!
Well, that's one thing in your formula's favor. You didn't write the formula with the quantity (4000/(4000+h))^2, you only wrote it with the denominator squared. So my answer below is wrong.

That might fix it. The formula would be Newton's gravitation formula equated with the weight force:
$F = \frac{GmM}{(R + h)^2} = mg$
where G is the Universal gravitation constant, m is the mass of the object, M is the mass of the planet, R is the radius of the planet, h is the height above the planet's surface, and g is the acceleration due to gravity at the height h.

Now, weight is defined as w = mg, so
$w = \frac{GmM}{(R + h)^2}$

Now, a couple of things are happening here. The object's mass is being weighed in lbs so on the RHS we have a mass m (a constant) in lbs, but we are looking for the weight (which depends on h) in lbs. This seriously messes with my head. The other thing is I have no idea what G is in the English system and I'm not about to try and figure it out.

With the change in the formula this may now be correct. I'm just not about to spend the time to prove it now.

-Dan

5. ## ok

Thanks for sharing all the valuable math data.

6. Originally Posted by topsquark
Well, that's one thing in your formula's favor. You didn't write the formula with the quantity (4000/(4000+h))^2, you only wrote it with the denominator squared. So my answer below is wrong.

That might fix it. The formula would be Newton's gravitation formula equated with the weight force:
$F = \frac{GmM}{(R + h)^2} = mg$
where G is the Universal gravitation constant, m is the mass of the object, M is the mass of the planet, R is the radius of the planet, h is the height above the planet's surface, and g is the acceleration due to gravity at the height h.

Now, weight is defined as w = mg, so
$w = \frac{GmM}{(R + h)^2}$

Now, a couple of things are happening here. The object's mass is being weighed in lbs so on the RHS we have a mass m (a constant) in lbs, but we are looking for the weight (which depends on h) in lbs. This seriously messes with my head. The other thing is I have no idea what G is in the English system and I'm not about to try and figure it out.

With the change in the formula this may now be correct. I'm just not about to spend the time to prove it now.

-Dan
You don't need to know $M_{\bigoplus}$ or $G$ to derive the correct relation here, just the knowlege that an inverse square law is involved. Then the ratio of gravitational forces on a body at a distance $R_{\bigoplus}+h$ to those at distance $R_{\bigoplus}$
is:

$\rho=\left[\frac{R_{\bigoplus}}{R_{\bigoplus}+h}\right]^2$,

so if $w(x)$ is the force at distance $x$ from the centre of the earth:

$w(R_{\bigoplus}+h)=w(R_{\bigoplus})\left[\frac{R_{\bigoplus}}{R_{\bigoplus}+h}\right]^2$

where $R_{\bigoplus}$ and $h$ are measured in the same units.

RonL

7. Originally Posted by CaptainBlack
You don't need to know $M_{\bigoplus}$ or $G$ to derive the correct relation here, just the knowlege that an inverse square law is involved. Then the ratio of gravitational forces on a body at a distance $R_{\bigoplus}+h$ to those at distance $R_{\bigoplus}$
is:

$\rho=\left[\frac{R_{\bigoplus}}{R_{\bigoplus}+h}\right]^2$,

so if $w(x)$ is the force at distance $x$ from the centre of the earth:

$w(R_{\bigoplus}+h)=w(R_{\bigoplus})\left[\frac{R_{\bigoplus}}{R_{\bigoplus}+h}\right]^2$

where $R_{\bigoplus}$ and $h$ are measured in the same units.

RonL
Clever. I hadn't thought to go that direction.

-Dan

W = Wi((4000)/(4000 + h))^2
where
W = weight at given height
Wi = weight at sea level (about 4000 miles from center of earth)
h - height in miles

If the initial weight is 120 pounds, and the elevation is 3 miles . . .

then,
W = 120((4000)/(4000 + 3))^2
W = 120(4000/4003)^2
W = 120 (0.99925)^2
W = 120 * .9985
W = 119.82 pounds (approximately)