(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 6.38 106 m.
(b) Find the satellite's orbital period.
Becasue of the inverse swquare law of gravity, the value of g at that orbital altitude will be 1/36 of the value at the earth's surface, which make is g/36. Thus
Centripetal acceleration =
V^2/(6*Re) = g/36
where Re is the radius of the Earth.
V^2 = Re*g/6 = 1.043*10^7
V = 3.32*10^3 m/s
For the orbit period T:
2 pi Re = V*T
T = 1.241*10^4 s = 207 minutes
To find the speed and orbital period of a satellite moving around the Earth in a circular orbit, you can use the following equations:
(a) Speed of the satellite:
The speed of a satellite moving in a circular orbit can be calculated using the formula:
v = √(G * M / r)
Where:
v = Speed of the satellite
G = Gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2)
M = Mass of the Earth (5.972 × 10^24 kg)
r = Radius of the satellite orbit
In this case, the radius of the satellite's orbit is given as 6 times the Earth's radius, which is 6.38 × 10^6 m. Therefore, r = 6 × 6.38 × 10^6 m.
Plugging the values into the equation, we get:
v = √((6.67430 × 10^-11 m^3 kg^-1 s^-2) * (5.972 × 10^24 kg) / (6 × 6.38 × 10^6 m))
Calculating the value of v will give you the speed of the satellite.
(b) Orbital period of the satellite:
The orbital period of a satellite can be calculated using the formula:
T = 2π * (r / v)
Where:
T = Orbital period of the satellite
r = Radius of the satellite orbit
v = Speed of the satellite (calculated in part (a)).
Plugging the values into the equation will give you the orbital period of the satellite.