a geostationary satellite for communications seems to be in a fixed spot above the equator because it has the same angular velocity as the earth.
show that if it goes round once a day its angular velocity w is a little over 7*10^-5 rad/s.?
Abebe
To show that the angular velocity (ω) of a geostationary satellite for communications is a little over 7x10^-5 rad/s, we need to understand the relationship between its orbital period and angular velocity.
1. The orbital period (T) is the time it takes for a satellite to complete one orbit around the Earth. In the case of a geostationary satellite, its orbital period is equal to one day (24 hours).
2. The angular velocity (ω) is the rate at which the satellite moves around its orbit, calculated in radians per second. It represents the angle covered per unit time.
Now, we can use the formula relating angular velocity, orbital period, and angle covered:
ω = 2π / T
where:
ω = angular velocity in rad/s
T = orbital period in seconds
π = pi (approximately 3.14159)
Substituting the orbital period for one day (24 hours or 86,400 seconds) into the formula:
ω = 2π / (24 x 3600) [since there are 24 hours in a day and 3600 seconds in an hour]
Simplifying, we get:
ω = π / (12 x 3600)
Calculating this expression:
ω ≈ 2.6179938 x 10^-5 rad/s
Therefore, the angular velocity of a geostationary satellite for communications is approximately 2.62 x 10^-5 rad/s, which is a little over 7x10^-5 rad/s.
Please note that this calculation assumes only the Earth's rotation period and does not consider any other factors like gravitational effects or variations in satellite orbits.
To show that the angular velocity of a geostationary satellite is a little over 7*10^-5 rad/s, we can use the relation between angular velocity, rotational period, and angular displacement.
Angular velocity (ω) is defined as the change in angular displacement per unit of time. It can be calculated using the formula:
ω = Δθ / Δt
Where:
ω is the angular velocity,
Δθ is the change in angular displacement, and
Δt is the change in time.
For a geostationary satellite, it orbits the Earth once in 24 hours, which is equivalent to 86,400 seconds (since there are 24 hours * 60 minutes * 60 seconds in a day). Therefore, the time it takes for the satellite to complete one full revolution is Δt = 86,400 seconds.
Since the satellite completes a full circular orbit, the change in angular displacement, Δθ, is 2π radians (360 degrees).
Substituting these values into the formula, we can solve for the angular velocity:
ω = 2π rad / 86,400 s
Simplifying this expression, we get:
ω ≈ 7.27 * 10^-5 rad/s
So, the angular velocity (w) of a geostationary satellite is a little over 7*10^-5 rad/s.