A 2000 kg satellite orbits the earth at a height of 300

km. What is the speed of the satellite and its period?
Take
, Mass of the earth is
kg

A 2000 kg satellite orbits the earth at a height of 300 km. What is the speed of the satellite and its period? Take

G 67 0 Nm kg , Mass of the earth is 5 98 0 kg

To calculate the speed of the satellite, we can use the formula for the orbital velocity:

v = sqrt(G * M / r)

where:
- v is the orbital velocity
- G is the gravitational constant (approximately 6.67 x 10^-11 N m^2/kg^2)
- M is the mass of the Earth
- r is the distance between the center of the Earth and the satellite

First, let's convert the height of the orbit from kilometers to meters:

height = 300 km = 300,000 m

To calculate the distance between the center of the Earth and the satellite, we need to add the radius of the Earth to the height of the orbit:

radius of the Earth = 6,371 km = 6,371,000 m

r = height + radius of the Earth = 300,000 m + 6,371,000 m = 6,671,000 m

Now, we can calculate the orbital velocity:

v = sqrt(G * M / r)

Substituting the given values:

v = sqrt((6.67 x 10^-11 N m^2/kg^2) * (5.97 x 10^24 kg) / 6,671,000 m)

Calculating this expression, we find that the speed of the satellite is approximately 7,660 m/s.

Now let's calculate the period (T) of the satellite. The period is the time it takes for the satellite to complete one orbit around the Earth. It can be calculated using the formula:

T = 2πr / v

Substituting the given values:

T = 2π * 6,671,000 m / 7,660 m/s

Simplifying this expression, we find that the period of the satellite is approximately 5,238 seconds.

So, the speed of the satellite is approximately 7,660 m/s and its period is approximately 5,238 seconds.