(a) Find the speed of a satellite moving around the earth in a circular orbit that has a radius equal to six times the earth's radius of 6.38 x 10^6 m.

(b) Find the satellite's orbital period.

centripetalforce=gravity force

v^2/r=9.8/(6)^2

V= sqrt (6*re*9.8/36)

checkthat.

Period? Well, T=distance/velocity=2PI*6Re/above V

That will give the period in seconds, you probably want to convert it.

(a) centrieptal acceleration rate =

V^2/r = GM/r^2

where M is the mass of the earth and r = 6 Re.
Re is the radius of the earth that you were given
G is the universal constant of gravity.

Solve for V

V^2/(6 Re) = GM/(6Re)^2

You can save yourself the trouble of looking up M and G by using the relationship
g = GM/Re^2. Then
6 V^2 /Re = GM/Re^2 = g
V^2 = (Re*g/6)
V = 3230 m/s

(b) V*(period) = 2 pi R = 12 pi Re

Solve for the period

Vc = sqrt(µ/r) where µ = 3.9863x10^14 and r = 6378km.

At 38,268km,
Vc = sqrt[3.9863x10^14/6x6,378,000] =

The period derives from
T = 2(Pi)sqrt[r^3/µ]

To find the answers for both (a) and (b), we need to use the laws of circular motion and gravitational force.

(a) The speed of a satellite moving in a circular orbit can be found using the formula:

v = √(G * M / r)

where:
v = speed of the satellite
G = gravitational constant, approximately 6.67430 x 10^(-11) m^3/(kg*s^2)
M = mass of the Earth, approximately 5.9722 x 10^24 kg
r = radius of the circular orbit

First, we need to calculate the radius of the circular orbit, which is given as six times the radius of the Earth:

radius = 6 * (6.38 x 10^6) m
= 3.828 x 10^7 m

Now, we can substitute the values into the formula:

v = √(6.67430 x 10^(-11) * 5.9722 x 10^24 / 3.828 x 10^7)

Simplifying this expression will give us the speed of the satellite.

(b) The orbital period of a satellite can be found using the formula:

T = 2 * π * r / v

where:
T = orbital period
π = pi, approximately 3.14159
r = radius of the circular orbit
v = speed of the satellite

Using the calculated radius and speed from part (a), we can substitute them into the formula to find the orbital period.