volume_mute
Your weight on earth and beyond it
publish date: 2024/03/22 23:38:59.505572 UTC
volume_muteYour distance to the center of the earth is about 4000 miles when I am on the surface. If you go to a height of 4000 miles above the surface of the earth, the force of gravity (your weight) then becomes what fraction of your present weight?
\(\frac12\)
\(\frac13\)
\(\frac14\)
Correct Answer
\(\frac14\)
Explanation
Let's denote the following:
\(d_1\) = Initial distance from the center of the Earth (4000 miles)
\(d_2\) = New distance from the center of the Earth (8000 miles, considering the 4000 miles height increase)
\(F_1\) = Initial force of gravity (weight) on the surface of the Earth
\(F_2\) = Force of gravity (weight) at a height of 4000 miles above the surface
According to Newton's law of universal gravitation:
\(F_1 = \frac{M \cdot m}{{d_1}^2}\)
\(F_2 = \frac{M \cdot m}{{d_2}^2}\)
Where:
- \(m\) is the mass of the object (in this case, the person),
- \(M\) is the mass of the Earth.
Now, to find the fraction of the present weight \( F_2 / F_1\), we divide \(F_2\) by \(F_1\):
\(\frac{F_2}{F_1} = \frac{\frac{M \cdot m}{{d_2}^2}} {\frac{M \cdot m}{{d_1}^2}}\)
\(\frac{F_2}{F_1} = \frac{{d_1}^2}{{d_2}^2}\)
\(\frac{F_2}{F_1} = \frac{4000^2}{8000^2} = \frac14\)
So, when you are 4000 miles above the surface of the Earth, your weight becomes \(\frac14\) of your present weight.
Reference
Basic Physics: A Self-Teaching Guide