Question

A space probe is launched from Earth headed for deep space. At a distance of 10,000 miles from Earth's center, the gravitational force on it is 520 lb. What is the size of the force when it is at each of the following distances from Earth's center? (a) 20,000 miles lb (b) 30,000 miles lb (c) 100,000 miles lb

Answer 1

(a) 130 lb

(b) 57.78 lb

(c) 5.20 lb

Step-by-step explanation:

The gravitational force acting on between 2 bodies is inversely proportional to the square of the distance between them.

If
`F_(g)` is gravitational force and
`r` is the distance, then

`F_(g)=(K)/(r^(2))`

Where,
`K` is a constant of proportionality.

Given:

`F_(g)=520\textrm { lb},r=10,000\textrm{ mi}`

∴
`F_(g)=(K)/(r^(2))`

`520=(K)/((10000)^(2))`

`K=520* 10^(8)` lb/mi²

(a) When
`K=520* 10^(8),r=20,000\textrm{ mi}`

∴
`F_(g)=(K)/(r^(2))\\F_(g)=(520* 10^(8))/((20000)^(2))\\F_(g)=(520* 10^(8))/(4* 10^(8))\\F_(g)=130 \textrm{ lb}`

Therefore, the gravitational force is 130 lb when the distance is 20,000 miles.

(b) When
`K=520* 10^(8),r=30,000\textrm{ mi}`

∴
`F_(g)=(K)/(r^(2))\\F_(g)=(520* 10^(8))/((30000)^(2))\\F_(g)=(520* 10^(8))/(9* 10^(8))\\F_(g)=57.78 \textrm{ lb}`

Therefore, the gravitational force is 57.78 lb when the distance is 30,000 miles.

(c) When
`K=520* 10^(8),r=100,000\textrm{ mi}`

∴
`F_(g)=(K)/(r^(2))\\F_(g)=(520* 10^(8))/((100000)^(2))\\F_(g)=(520* 10^(8))/(10* 10^(8))\\F_(g)=5.20 \textrm{ lb}`

Therefore, the gravitational force is 5.20 lb when the distance is 100,000 miles.