Question

An earth satellite moves in a circular orbit with an orbital speed of 6200 .

Find the time of one revolution of the satellite. in seconds
Find the radial acceleration of the satellite in its orbit. in M/s2

Answer 1

The time of one revolution (period) and the radial acceleration of an Earth satellite cannot be determined with only the given orbital speed of 6200 m/s; orbit radius information is required as well.

Step-by-step explanation:

To find the time of one revolution of the satellite, one would need the orbit circumference and the orbital speed. Since the question does not provide the orbit radius, but only gives the orbital speed of 6200 m/s, we cannot calculate the period without more information.

To find the radial acceleration (often referred to as centripetal acceleration) of the satellite, you would typically use the formula ar = v2 / r, where v is the orbital speed and r is the radius of the orbit. Again, without the orbit radius, this calculation cannot be completed. For an Earth satellite to remain in a circular orbit, the centripetal acceleration must be provided by the gravitational force exerted by Earth, such that Fg = m * ar.

Answer 2

The radial acceleration of the satellite in its orbit is 3.70 m/s²

What is the radial acceleration of the satellite in its orbit?

Orbital speed of satellite, v = 6200 m/s

`v = \sqrt{ ( GM_(E) )/(r) }`

The radius, r

`r = \frac{ GM_(E) }{ {v}^(?) }`

`= \frac{(6.673 * {10}^( - 11) {Nm}^(2) {kg}^(2))(5.97 * {10}^(24)kg) }{ {(6200m/s)}^(2) }`

`r = 1.04 * {10}^(7) m`

`T = (2\pi r)/(v)`

`= \frac{2\pi(1.04 * {10}^(7)m) }{ {6200m/s} }`

`T = 1.05 * {10}^(4)s`

The radial acceleration of the satellite in its orbit

`a_(r) = \frac{ {v}^(2) }{r}`

`= \frac{ ({6200m/s)}^(2) }{1.04 * {10}^(7) m}`

`a_(r) = 3.70 {m/s}^(2)`