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.A) show that if it goes round once a day,its angular velocity is a little over 7×10 race to negative 5 rad per second

w=2PI/time=2PI/1day*1day/24hrs*1hr/3600sec= 2PI/(24*3600)=

= 7.27220522e-5 rad/sec

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Well, you see, the geostationary satellite is like that one friend who always manages to stay in the same spot at a party, never missing any action. It's a pro at synchronizing with the Earth's rotation, which is impressive.

To calculate its angular velocity, we need to know the time it takes for the satellite to complete one orbit around the Earth, which is one day. We also need to know the circumference of the Earth, which is approximately 40,075 kilometers.

Angular velocity is defined as the angle covered per unit time. In this case, we need to know the angle covered in one day by the satellite. Since it's a complete orbit, we know that the angle covered is 360 degrees or 2π radians.

Now, let's do some math. We'll start by converting the distance traveled by the satellite in one day to radians:

Circumference of Earth = 40,075 km
Distance traveled by satellite in one day = 40,075 km

To convert this distance to radians, we'll use the formula:

Distance in radians = Distance in kilometers / Radius of Earth

Radius of Earth is approximately 6,371 km:

Distance in radians = 40,075 km / 6,371 km

So, the distance traveled by the satellite in radians in one day is approximately 6.28 radians.

Now we'll calculate the angular velocity:

Angular velocity = Distance in radians / Time taken

Time taken = 24 hours = 24 * 60 * 60 seconds

Angular velocity = 6.28 radians / (24 * 60 * 60 seconds)

After a bit of calculation, we get the value:

Angular velocity ≈ 7.27 x 10^(-5) radians per second

Voilà! That's a little over 7 times 10 raised to the power of negative 5 radians per second. Quite the mouthful, isn't it? But don't worry, this satellite is a slacker compared to some of its space buddies who zip around the universe even faster.

So, that's how this geostationary satellite manages to keep up with Earth's rotation. It's the perfect party companion, always in the right spot at the right time!

To show that the angular velocity of a geostationary satellite is a little over 7×10^-5 rad per second, we need to consider the relationship between angular velocity, time period, and angle moved.

Angular velocity (ω) is defined as the change in angle (θ) divided by the change in time (t). Mathematically, it can be expressed as ω = θ / t.

Given that the geostationary satellite completes one full revolution in one day, we can determine the time period (t) as 24 hours or 24 × 60 × 60 seconds. Therefore, t = 86400 seconds.

Since the satellite completes one revolution in a day, the angle moved by the satellite (θ) will be 2π radians (since one complete revolution is equivalent to 2π radians).

Plugging in the values into the formula for angular velocity, we get:

ω = θ / t
= 2π / (24 × 60 × 60)
= (2π) / 86400

Calculating the approximate numerical value, we have:

ω ≈ (2 × 3.14) / 86400
≈ 0.000072722 rad per second
≈ 7.27 × 10^-5 rad per second

Therefore, we can conclude that if a geostationary satellite goes round once a day, its angular velocity is approximately 7.27 × 10^-5 rad per second, which is a little over 7×10^-5 rad per second.