A satellite of mass 225 kg is launched from a site on Earth's equator into an orbit at 200 km above the surface of Earth. (The mass of the Earth is 5.98 1024 kg, and the radius of the Earth is 6.38 103 km.) (a) Assuming a circular orbit, what is the orbital period of this satellite?

To calculate the orbital period of a satellite in a circular orbit, we can use Kepler's third law of planetary motion, which states:

T^2 = (4π^2 * r^3) / (G * M)

Where:
T = Orbital period (in seconds or any other time unit)
r = Distance between the center of the Earth and the satellite (in meters or any other distance unit)
G = Universal gravitational constant (6.67430 × 10^-11 N(m/kg)^2)
M = Mass of the Earth (in kilograms or any other mass unit)

First, we need to convert the given distance to meters:
Distance = 200 km = 200,000 meters

Second, we need to convert the mass of the Earth to kilograms:
Mass of Earth = 5.98 * 10^24 kg

Now we can substitute these values into the equation and solve for T:

T^2 = (4π^2 * r^3) / (G * M)

T^2 = (4 * 3.14159^2 * (200,000)^3) / (6.67430 × 10^-11 * 5.98 * 10^24)

T^2 = 1.2814 * 10^20 / (3.9557 * 10^14)

T^2 = 324.205

Taking the square root of both sides, we get:

T = √324.205

T ≈ 17.987 seconds

Therefore, the orbital period of the satellite is approximately 17.987 seconds.

To determine the orbital period of a satellite, you can use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period (T) is proportional to the cube of the average distance between the satellite and the center of the Earth (r).

The average distance (r) between the satellite and the center of the Earth is the sum of the radius of the Earth (R) and the altitude of the satellite (h):
r = R + h

First, we need to convert the radius of the Earth from km to meters:
R = 6.38 × 10^3 km * 10^3 m/km = 6.38 × 10^6 m

Next, we need to convert the altitude of the satellite from km to meters:
h = 200 km * 10^3 m/km = 2 × 10^5 m

Now we can substitute the values into the equation for r:
r = 6.38 × 10^6 m + 2 × 10^5 m = 6.58 × 10^6 m

Next, we need to calculate the gravitational parameter (μ) of the Earth, which is the product of the gravitational constant (G) and the mass of the Earth (M):
μ = G * M

The gravitational constant is approximately 6.674 × 10^(-11) N·m^2/kg^2, and the mass of the Earth is 5.98 × 10^24 kg.

Substitute the values into the equation for μ:
μ = 6.674 × 10^(-11) N·m^2/kg^2 * 5.98 × 10^24 kg = 3.986 × 10^14 N·m^2/kg

Now we can calculate the orbital period (T) using Kepler's Third Law:
T^2 = (4 * π^2 * r^3) / μ

Substitute the values into the equation:
T^2 = (4 * π^2 * (6.58 × 10^6 m)^3) / (3.986 × 10^14 N·m^2/kg)
T^2 = 3.16 × 10^22

To solve for T, take the square root of both sides of the equation:
T = √(3.16 × 10^22)
T ≈ 5.62 × 10^11 seconds

Therefore, the orbital period of the satellite is approximately 5.62 × 10^11 seconds.

F = m a

G m M/r^2 = m V^2/r
m, the mass of the satellite, is irrelevant, cancels.

G M = V^2 r

distance around circumference = 2 pi r
so t = 2 pi r / V
or V = 2 pi r/t
so
G M = (2 pi)^2 (r^3/t^2)
or
t^2 = (2 pi)^2 r^3/ (G M)
here
G = 6.67*10^=11
M = 5.98 *10^24 Kg
r = 6.38*10^6 + .2*10^6
= 6.58*10^6 meters
so
t^2 = (2 pi)2 *(6.58)^3*(10)^18/(6.57*5.98*10^13)
in seconds