Question

1. A satellite (mass = 4.44 x 109 kg) travels in orbit around the Earth at a distance of 1.9 x 10'm above

Earth's surface. What is the force of gravitational attraction between the satellite and Earth?

Answer 1

The force of gravitational attraction between the satellite and Earth is
`2.587* 10^(10)` newtons.

Step-by-step explanation:

Statement is incorrect. Correct statement is:

A satellite (
`m = 4.44* 10^(9)\,kg`) travels in orbit around the Earth at a distance of
`1.9* 10^(6)\,m` above Earth's surface. What is the force of gravitational attraction between the satellite and Earth?

The gravitational force experimented by the satellite (
`F`), in newtons, is calculated by Newton's Law of Gravitation, whose equation is defined by following formula:

`F = (G\cdot m \cdot M)/(R^(2))` (1)

Where:

`G` - Gravitational constant, in cubic meters per kilogram-square second.

`m` - Mass of the satellite, in kilograms.

`M` - Mass of the Earth, in kilograms.

`R` - Distance of the satellite with respect to the center of the Earth, measured in meters.

If we know that
`G = 6.674* 10^(-11)\,(m^(3))/(kg\cdot s^(2))`,
`m = 4.44* 10^(9)\,kg`,
`M = 5.972* 10^(24)\,kg` and
`R = 8.271* 10^(6)\,m`, then the force of gravitational attraction between the satellite and Earth is:

`F = 2.587* 10^(10)\,N`

The force of gravitational attraction between the satellite and Earth is
`2.587* 10^(10)` newtons.