Grade 12 Physics: Satellite Question | Distance & Speed

[AngularVelocity]

Hello I am in Grade 12 Physics and I would like some help on this problem as soon as possible. Thanks.

A satellite travels directly above the equator, in the same direction as the Earth's spin, and completes two orbits of the Earth in one day. Determine the distance above the Earth's surface, and the speed, of the satellite.

 

[denverdoc]

Welcome to the forum. We ask that you at least try to post some ideas about solution to the problem. Do you know equation for uniform circular motion that relates velocity and distance?

 

[m2003]

Hello, thank you for reaching out for help with this physics problem. To determine the distance above the Earth's surface and the speed of the satellite, we need to use some key equations and principles from physics.

First, let's define some variables. Let's call the distance above the Earth's surface "h" and the speed of the satellite "v". We also know that the satellite completes two orbits in one day, which means it travels a distance equal to the circumference of the Earth twice in one day.

Using the equation for circumference, we can say that:

2πr = 2h

Where r is the radius of the Earth. We can rearrange this equation to solve for h:

h = πr

Next, we can use the equation for the speed of an object in circular motion:

v = 2πr/T

Where T is the time it takes to complete one orbit. Since the satellite completes two orbits in one day, T is equal to 12 hours or 43200 seconds. Substituting this into the equation, we get:

v = 2πr/43200

Now, we can substitute our value for h (πr) into this equation:

v = 2(πr)/43200

Simplifying, we get:

v = πr/21600

We can also use the equation for the distance an object travels in circular motion:

d = vt

Where d is the distance, v is the speed, and t is the time. In this case, the distance is equal to the circumference of the Earth, which we know is 2πr. Substituting this into the equation, we get:

2πr = v(43200)

Simplifying, we get:

r = v(43200)/2π

Now, we can substitute this value for r into our equation for h:

h = π(v(43200)/2π)

Simplifying, we get:

h = 21600v

Therefore, we have two equations that relate h and v:

h = πr

h = 21600v

We can solve for h in the second equation and substitute it into the first equation to get:

πr = 21600v

πr = 21600(πr)

Simplifying, we get:

r = 21600v

Now, we can substitute this value for r into our equation for v: