Question

A satellite moves on a circular earth orbit that has a radius of 6,758,998 m. A model airplane is flying on a 10 m guideline in a horizontal circle. The guideline is parallel to the ground. Find the speed of the plane such that the plane and the satellite have the same centripetal acceleration.

Answer 1

v = 0.1068 m / s

Step-by-step explanation:

To find the speed of the satellite we use Newton's second law where the force is the universal law of gravitation

F = ma

F =
`G (m M)/(r^2)`

acceleration is centripetal

a = v² / r

we substitute

`G (m M)/(r^2) = m (v^2)/(r)`

v² =
`G (M)/(r)`

The radius of the orbit is given we will assume that this radius is half from the center of the earth

we substitute

v² = 6.67 10⁻¹¹ 5.98 10²⁴/6758998

v =
`√(59.013 \ 10^6)`

v = 7.68 10³ m / s

The centripetal acceleration is

a = v² / r

a = 7.68 10³/6758998

a = 1.14 10⁻³ m / s²

For the airplane we use the definition of centripetal acceleration

a = v² / r

v =
`√(a \ r )`

let's calculate

v =
`\sqrt{1.14 \ 10^(-3) \ 10}`

v = 0.1068 m / s