Question

Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of your current weight. Show all work leading to your answer OR describe your solution using 3 -4 complete sentences.

Answer 1

The weight of an object is due to the attractive force of gravity.
The gravitational attraction of the earth decreases as the inverse square of the distance from the centre, for points outside the earth.

Answer 2

So what would be the equation for that? @ProfBrainstorm

Answer 3

\[F=\frac{ k }{ r^2 }\] where F is the force or weight, k is a constant and r is the distance from the centre of the Earth

Answer 4

let \[g _{1}=g \], g is your acceleration due to gravity on earth
and let \[r _{1}=6400km\]
\[F \alpha \frac{ 1 }{ r^{2} }\] \[mg \alpha \frac{ 1 }{ r^{2} }\] As the mass is not changing,
\[g \alpha \frac{ 1 }{ r^2 }\]
Therefore \[\frac{ g _{1} }{ g _{2} }=\frac{ \frac{ 1 }{ r _{1}^{2} } }{ \frac{ 1 }{ r _{2}^{2} } }=\frac{ r _{2}^{2} }{ r _{1}^{2} }\]
Given \[g _{2}=\frac{ g }{ 16 }\]
\[r _{2}^{2}=(\frac{ g_{1} }{ g_{2} })*r_{1}^{2}\]
\[r_{2}^{2}=(\frac{ \frac{ g }{ 1 }}{ \frac{ g }{ 16 } })*(6400)^{2}\]
\[r_{2}^{2}=16*6400^{2}=4^{2}*6400^{2}\]
\[r_{2}=4*6400=25,600km\]

Answer 5

Thank you!! can you guys help me with another one?