Answer: 200 lb
Explanation:
The rest of the question is written below:
What is the size of the force when it is at 30,000 miles from the Earth's center?
According to Newton's Universal Law of Gravitation:
[tex]F=G\frac{Mm}{r^2}[/tex] (1)
Where:
[tex]F[/tex] is the gravitational force
[tex]G[/tex]is the gravitational constant
[tex]M[/tex] is the mass of the Earth
[tex]m[/tex] is the mass of the probe
[tex]r[/tex] is the distance between the Earth and the probe
In this case we have:
[tex]F_{1}=800 lb[/tex] gravitational force when the distance is [tex]r_{1}=15000 mi[/tex] and we have to find [tex]F_{2}[/tex] gravitational force when the distance is [tex]r_{2}=30000 mi[/tex].
Then:
[tex]F_{1}=G\frac{Mm}{r_{1}^2}[/tex] (2)
[tex]F_{2}=G\frac{Mm}{r_{2}^2}[/tex] (3)
Dividing (2) by (3):
[tex]\frac{F_{1}}{F_{2}}=\frac{G\frac{Mm}{r_{1}^2}}{G\frac{Mm}{r_{2}^2}}[/tex] (4)
Simplifying:
[tex]\frac{F_{1}}{F_{2}}=\frac{r_{2}^2}{r_{1}^2}[/tex] (5)
Finding [tex]F_{2}[/tex]:
[tex]F_{2}=F_{1}\frac{r_{2}^2}{r_{1}^2}[/tex] (6)
[tex]F_{2}=800 lb\frac{(15000 mi)^2}{(30000 mi)^2}[/tex] (7)
Finally:
[tex]F_{2}=200 lb[/tex]
