# Questions About Satellites Orbiting Earth

• mrsvonnegut
In summary, to put a satellite with a mass of 250 kg in orbit at an altitude of 4.0 x 106 m, the total energy needed would be 6007709504 J. To calculate the period of the satellite at that altitude, the formula T=2π√(r3/GM) can be used. To put the satellite into a geosynchronous orbit, the additional energy needed would be the kinetic energy of the velocity at Earth's period, added to the potential energy needed to reach the geosynchronous orbit. The PE for reaching the geosynchronous orbit can be calculated using the formula PE=G(mM/r).
mrsvonnegut

## Homework Statement

I have a satellite with a mass of 250 kg, which I'm launching from Earth. This neglects air resistance/rotation of the planet etc.
• a. How much energy is needed to put the satellite in orbit at an altitude of 4.0 x 106 m?
• b. What would be the period of the satellite at that altitude?
• c. How much additional energy would be needed to put the satellite into a geosynchronous orbit?

Period=2pir/T
PE=G(mM/r)
KE=(1/2)(m)(v^2)

## The Attempt at a Solution

This is my first day doing orbital problems, so I'm really having trouble getting started! For a, I thought maybe I should calculate velocity then the kinetic energy using 1/2mv^2? And then add it to the potential energy of the satellite when it's orbiting, using PE=mgh?For b, I know a formula to find the period. For c, I've never seen a problem like that, but I assume I would put 24 hours into the period equation and start from there? Thank you in advance!

Last edited:
For part a, you just need to get the PE from the surface of the Earth to that altitude.

Okay! I did:
PEi=G([(250 kg)(5.97 x 10^24 kg)]/[6378100 m])
PEf=G([(250 kg)(5.97 x 10^24 kg)]/[6378100 m + 4.0 x 10^6 m])
The difference between the two was 6007709504 J.
(5.97 x 10^24 kg is what I used as the mass of the earth, and 6378100 m is what I used as the radius of the earth).
Does that look okay?

mrsvonnegut said:
Okay! I did:
PEi=G([(250 kg)(5.97 x 10^24 kg)]/[6378100 m])
PEf=G([(250 kg)(5.97 x 10^24 kg)]/[6378100 m + 4.0 x 10^6 m])
The difference between the two was 6007709504 J.
(5.97 x 10^24 kg is what I used as the mass of the earth, and 6378100 m is what I used as the radius of the earth).
Does that look okay?

I think that should be correct.

For the second part, use the fact that gravitation force of attraction between the Earth and the satellite = centripetal force on the satellite to get the period.

Question for pondering: Do you want to take into account the fact that the satellite, when it's sitting on the launching pad, already has a speed around the Earth's center due to the rotation of the Earth?

gneill said:
Question for pondering: Do you want to take into account the fact that the satellite, when it's sitting on the launching pad, already has a speed around the Earth's center due to the rotation of the Earth?

Nope! My teacher said we're not considering it.

Okay, you've worked out the energy that you have to supply in order to climb to the appropriate height above the Earth, now you need to impart an orbital velocity (KE) in order that it can stay there. What's the required velocity for a circular orbit at a given orbital radius r?

I'm not sure how to show the gravitational force between Earth and the satellite--should I use mg?

Also, one of my friends told me just to calculate the PE of the satellite above Earth as PE=G([250 kg)(5.97 x 10^24)]/(4.0 x 10^6]), which gives a total of 24887437.5 J, which is a different answer. How can I tell which is right?

mg = m(v^2/r), but should I use regular gravity?

mrsvonnegut said:
I'm not sure how to show the gravitational force between Earth and the satellite--should I use mg?

Also, one of my friends told me just to calculate the PE of the satellite above Earth as PE=G([250 kg)(5.97 x 10^24)]/(4.0 x 10^6]), which gives a total of 24887437.5 J, which is a different answer. How can I tell which is right?

Your friend is confused about the distinction between altitude (above Earth's surface) and orbital radius (around Earth's center).

mrsvonnegut said:
mg = m(v^2/r), but should I use regular gravity?

'mg' only applies near the Earth's surface. Use the full Newtonian gravitation formula for the force (and the PE).

GmM/d^2 = m(v^2/r)?

*r^2

mrsvonnegut said:
GmM/d^2 = m(v^2/r)?

In this case d is the same as r, so you can get v which will help you to get ω.

Your first equation in the relevant equations should be

T = 2π/ω

What is ω? I don't think I've learned it yet. Is it a symbol?

mrsvonnegut said:
What is ω? I don't think I've learned it yet. Is it a symbol?

Angular velocity which is found in circular motion. It has a simple relation to v and r.

Oh--we haven't done that in my class! Is there a way to avoid it?

mrsvonnegut said:
Oh--we haven't done that in my class! Is there a way to avoid it?

Well usually when you learn circular motion and centripetal force (mv2/r), you would learn about angular velocity and how it relates to velocity and radius.

Are you given the formula for kepler's third law where T2∝r3 ?

Yes, I do have that!

mrsvonnegut said:
Yes, I do have that!

ah in that case you can rewrite T2∝r3 as T2=Kr3, your notes should have what is needed to calculate the constant K.

rock.freak667 said:
ah in that case you can rewrite T2∝r3 as T2=Kr3, your notes should have what is needed to calculate the constant K.

My teacher provides a way to find the period when I have the velocity, which I know from our earlier work! So for part c, should I substitute Earth's period, then find the new velocity, then find the new KE, and find the difference between the KE's?

mrsvonnegut said:
My teacher provides a way to find the period when I have the velocity, which I know from our earlier work!

Oh in that case, then find 'v' using the earlier way.

mrsvonnegut said:
So for part c, should I substitute Earth's period, then find the new velocity, then find the new KE, and find the difference between the KE's?

I think once you find the KE, you would add that to the PE to get the total energy required. The PE you calculated before is how much energy is required just to get it at that height.

Got it--so the KE of the v of Earth's period (to move it upward), plus my earlier work to put it in orbit. THANK YOU SO MUCH! You were so helpful and patient!

## 1. What are satellites and how do they orbit the Earth?

Satellites are objects that are launched into space and orbit around the Earth. They are typically used for communication, navigation, and scientific research. Satellites orbit the Earth due to the gravitational pull of the Earth. They travel at a high speed and their orbit is maintained by the balance between their velocity and the pull of gravity.

## 2. How many satellites are currently orbiting the Earth?

As of 2021, there are over 2,700 operational satellites orbiting the Earth. This number does not include inactive or non-functioning satellites, which adds up to thousands more.

## 3. What is the purpose of having multiple satellites in orbit?

Having multiple satellites in orbit allows for better coverage and communication. Since satellites have a limited coverage area, having multiple satellites ensures that there is always a satellite within range for communication and data transmission. It also allows for better accuracy in navigation and data collection.

## 4. How are satellites launched into orbit?

Satellites are typically launched into orbit using rockets. These rockets are designed to carry the satellite into space and then release it at a specific altitude and velocity for it to enter orbit. Some satellites are also launched using specialized vehicles such as space shuttles.

## 5. What happens to satellites once they reach the end of their lifespan?

Once a satellite reaches the end of its lifespan, it can either be deorbited and burned up in the Earth's atmosphere, or it can be moved to a higher orbit known as a "graveyard orbit". This is to prevent the satellite from becoming space debris and potentially colliding with other satellites in orbit.

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