Question

The value of the gravitational acceleration g decreases with elevation from 9.807 m/s2 at the sea level to 9.767 m/s2 at an altitude of 13000 m, where large passenger planes cruise. Determine the percentage reduction in the weight of an airplane cruising at 13000 m relative to its weight at sea level.

Answer 1

Answer: 0.41%

Explanation:

If we want to calculate the percentage decrease (the reduction) in weight of an airplane, we can use the given values of gravity, because the weight is directly proportional to the gravity acceleration.

Now the formula for the percentage decrease for ths case is:

$\% weightdecrease=\frac{g_{sea}-g_{13000m}}{g_{sea}}(100)$

Where:

$g_{sea}=9.807 m/s^{2}$ is the gravity at sea level

$g_{13000m}=9.767 m/s^{2}$ is the gravity at 13000 m

$\% weightdecrease=\frac{9.807 m/s^{2} - 9.767 m/s^{2}}{9.807 m/s^{2}}(100)$

$\% weightdecrease=0.00407 (100)$

$\% weightdecrease=0.407 \% \approx 0.41 \%$ This is the percentage reduction in the weight of an airplane