Respuesta :
Answer:
(a) 130 lb
(b) 57.78 lb
(c) 5.20 lb
Explanation:
The gravitational force acting on between 2 bodies is inversely proportional to the square of the distance between them.
If [tex]F_{g}[/tex] is gravitational force and [tex]r[/tex] is the distance, then
[tex]F_{g}=\frac{K}{r^{2}}[/tex]
Where, [tex]K[/tex] is a constant of proportionality.
Given:
[tex]F_{g}=520\textrm { lb},r=10,000\textrm{ mi}[/tex]
∴ [tex]F_{g}=\frac{K}{r^{2}}[/tex]
[tex]520=\frac{K}{(10000)^{2}}[/tex]
[tex]K=520\times 10^{8}[/tex] lb/mi²
(a) When [tex]K=520\times 10^{8},r=20,000\textrm{ mi}[/tex]
∴ [tex]F_{g}=\frac{K}{r^{2}}\\F_{g}=\frac{520\times 10^{8}}{(20000)^{2}}\\F_{g}=\frac{520\times 10^{8}}{4\times 10^{8}}\\F_{g}=130 \textrm{ lb}[/tex]
Therefore, the gravitational force is 130 lb when the distance is 20,000 miles.
(b) When [tex]K=520\times 10^{8},r=30,000\textrm{ mi}[/tex]
∴ [tex]F_{g}=\frac{K}{r^{2}}\\F_{g}=\frac{520\times 10^{8}}{(30000)^{2}}\\F_{g}=\frac{520\times 10^{8}}{9\times 10^{8}}\\F_{g}=57.78 \textrm{ lb}[/tex]
Therefore, the gravitational force is 57.78 lb when the distance is 30,000 miles.
(c) When [tex]K=520\times 10^{8},r=100,000\textrm{ mi}[/tex]
∴ [tex]F_{g}=\frac{K}{r^{2}}\\F_{g}=\frac{520\times 10^{8}}{(100000)^{2}}\\F_{g}=\frac{520\times 10^{8}}{10\times 10^{8}}\\F_{g}=5.20 \textrm{ lb}[/tex]
Therefore, the gravitational force is 5.20 lb when the distance is 100,000 miles.
