Question

A space probe is launched from earth, headed for deep space. At a distance of 15,000 miles from the Earth center, the gravitational force on it is 800 lb.

Answer 1

Answer: 200 lb

Step-by-step 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:

`F=G(Mm)/(r^2)` (1)

Where:

`F` is the gravitational force

`G`is the gravitational constant

`M` is the mass of the Earth

`m` is the mass of the probe

`r` is the distance between the Earth and the probe

In this case we have:

`F_(1)=800 lb` gravitational force when the distance is
`r_(1)=15000 mi` and we have to find
`F_(2)` gravitational force when the distance is
`r_(2)=30000 mi`.

Then:

`F_(1)=G(Mm)/(r_(1)^2)` (2)

`F_(2)=G(Mm)/(r_(2)^2)` (3)

Dividing (2) by (3):

`(F_(1))/(F_(2))=(G(Mm)/(r_(1)^2))/(G(Mm)/(r_(2)^2))` (4)

Simplifying:

`(F_(1))/(F_(2))=(r_(2)^2)/(r_(1)^2)` (5)

Finding
`F_(2)`:

`F_(2)=F_(1)(r_(2)^2)/(r_(1)^2)` (6)

`F_(2)=800 lb((15000 mi)^2)/((30000 mi)^2)` (7)

Finally:

`F_(2)=200 lb`