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# if I weigh 141.2lbs at sea level what will I weigh 29,000ft above sea level

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if I weigh 141.2lbs at sea level what will I weigh 29,000ft above sea level

Guest Feb 19, 2015

#1
+18843
+5

If I weigh 141.2lbs at sea level what will I weigh 29,000ft above sea level ?

29000 ft = 8839.2 m

Only Free-air correction:

Accounts for the 1/r^2 decrease in gravity with the distance from the center of the Earth:

$$g_1=\boxed{g_{(\text{ see level })}=\dfrac{GM_E}{R_E^2}}$$

and in 8839.2 m

$$g_2=\boxed{g_{(\ 29000\ ft \text{ above sea level})}=\dfrac{GM_E}{( R_E +8.8392\ km )^2}}$$

$$F_2 = F_1\cdot \dfrac{g_2}{g_1}\cdot \dfrac{\not{m}}{\not{m}}$$

$$\small{\text{  F_2 = F_1\cdot \dfrac{R_E^2} {( R_E +8.8392\ km )^2} =F_1\cdot \dfrac{6371^2} {( 6371 +8.8392\ km )^2}=F_1\cdot 0.99723094065  }}$$

So we have 141.2lbs * 0.99723094065 = 140.809008820 lbs

heureka  Feb 19, 2015
Sort:

#1
+18843
+5

If I weigh 141.2lbs at sea level what will I weigh 29,000ft above sea level ?

29000 ft = 8839.2 m

Only Free-air correction:

Accounts for the 1/r^2 decrease in gravity with the distance from the center of the Earth:

$$g_1=\boxed{g_{(\text{ see level })}=\dfrac{GM_E}{R_E^2}}$$

and in 8839.2 m

$$g_2=\boxed{g_{(\ 29000\ ft \text{ above sea level})}=\dfrac{GM_E}{( R_E +8.8392\ km )^2}}$$

$$F_2 = F_1\cdot \dfrac{g_2}{g_1}\cdot \dfrac{\not{m}}{\not{m}}$$

$$\small{\text{  F_2 = F_1\cdot \dfrac{R_E^2} {( R_E +8.8392\ km )^2} =F_1\cdot \dfrac{6371^2} {( 6371 +8.8392\ km )^2}=F_1\cdot 0.99723094065  }}$$

So we have 141.2lbs * 0.99723094065 = 140.809008820 lbs

heureka  Feb 19, 2015
#2
+91513
0

Thanks Heureka

I was thinking that I could go live on a mountain to lose some weight (Mt Everest is 29000 feet) but Heurika has determined that the diffenence will be 0.4 of a pound so it is probably not going to be worth the moving expense. How sad.

Melody  Feb 19, 2015

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