A satellite orbits the Earth in an orbits 750 km high. Find (*) the satellite's orbital speed and (**) period. (The mass and radius of the Earth are 5.98E + 24 kg and 6370000 m, respectively.)

the centripetal force (m v^2 / r)

is equal to the gravitational force
... G M m / r^2

v^2 = G M / r
... = 6.67E-11 * 5.98E24 / 7.12E6

the period is the speed divided into the orbital circumference

To find the orbital speed of a satellite, we can use the formula:

v = √(G * M / r)

Where:
v is the orbital speed,
G is the universal gravitational constant (6.67 x 10^-11 m^3 kg^-1 s^-2),
M is the mass of the Earth,
r is the distance from the center of the Earth to the satellite.

To find the period of the satellite, we can use the formula:

T = 2π * (√(r^3 / (G * M)))

Where:
T is the period,
G is the universal gravitational constant,
M is the mass of the Earth,
r is the distance from the center of the Earth to the satellite.

Let's calculate it step by step:

Step 1: Convert the height of the satellite into meters.
The height of the satellite is given as 750 km. We need to convert it to meters by multiplying it by 1000.
Height = 750 km * 1000 = 750000 m

Step 2: Calculate the distance from the center of the Earth to the satellite.
The distance from the center of the Earth to the satellite is the sum of the radius of the Earth and the height of the satellite.
r = 6370000 m + 750000 m = 7120000 m

Step 3: Calculate the orbital speed.
v = √(G * M / r)
Substituting the values, we have:
v = √((6.67 x 10^-11 m^3 kg^-1 s^-2) * (5.98 x 10^24 kg) / (7120000 m))
v ≈ 7644.621 m/s

Therefore, the satellite's orbital speed is approximately 7644.621 m/s.

Step 4: Calculate the period.
T = 2π * (√(r^3 / (G * M)))
Substituting the values, we have:
T = 2π * (√((7120000 m)^3 / ((6.67 x 10^-11 m^3 kg^-1 s^-2) * (5.98 x 10^24 kg))))
T ≈ 5787.714 s

Therefore, the satellite's period is approximately 5787.714 seconds.