Question

Find the percentage decrease in weight of body whem taken to a height of 32 km above the surface f earth. Radius of earth is given as 6400 km.

Answer 1

Question 37229

Find the percentage decrease in weight of body when taken to a height of 32 km above the surface of earth. Radius of earth is given as 6400 km.

Weight is defined as $P=mg=G\frac{Mm}{d^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $d$ is the distance from the object to the Earth’s center of the core.

When the object is at the Earth’s surface,

$P_{0}=G\frac{Mm}{R^{2}},$

where $R$ is the radius of the Earth.

When the object is taken to a height $h$ above the Earth’s surface,

$P_{h}=G\frac{Mm}{(R+h)^{2}}.$

The percentage difference of the new weight with respect to the original one is

$p.d.=\frac{P_{h}-P_{0}}{P_{0}}\times 100\%={\frac{\frac{1}{(R+h)^{2}}-\frac{1}{R^{2}}}{\frac{1}{R^{2}}}}\times 100\%=(\frac{R^{2}}{(R+h)^{2}}-1)\times 100\%.$

Using the numerical values, $R=6400$ km, and $h=$32 km, we substitute

$p.d.=(6400^{2}\over 6432^{2}-1)\times 100\%\approx-0.9925\%.$

Answer: the weight is about 0.9925% less.