Linear Velocity Calculation in Low Earth Orbit

In summary: It appears you are to give the answer in km/hr, in which case you would multiply the result you get from the formula I gave by 60. Your not going to get 28000 km/hr exactly, unless you use a value for the radius of the Earth constructed to give you that value for the linear speed. Using the value I cited, I get about 27663 km/hr.
  • #1
dtippitt
5
0
Can someone please check my work to this?
The reflecting telescope is deployed in low Earth orbit( 600km) with each orbit lasting about 95 min. use the linear velocity formula to solve the problem.

I did 300 * 95 min = 28500. Can someone check my work please? if someone could check it today hat would be great thanks.
 
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  • #2
dtippitt said:
Can someone please check my work to this?
The reflecting telescope is deployed in low Earth orbit( 600km) with each orbit lasting about 95 min. use the linear velocity formula to solve the problem.

I did 300 * 95 min = 28500. Can someone check my work please? if someone could check it today hat would be great thanks.

I would use:

\(\displaystyle v=r\omega=\left(r_E+600\right)\frac{2\pi}{95}\,\frac{\text{km}}{\text{min}}\)

where \(r_E\) is the radius of the Earth in km. Are you given a value for this that you are to use?
 
  • #3
MarkFL said:
I would use:

\(\displaystyle v=r\omega=\left(r_E+600\right)\frac{2\pi}{95}\,\frac{\text{km}}{\text{min}}\)

where \(r_E\) is the radius of the Earth in km. Are you given a value for this that you are to use?

the formula they gave me is v=r(radian symbol)/t
I think r stands for radius and t stands for time. I am not sure how to use this formula.
 
  • #4
dtippitt said:
the formula they gave me is v=r(radian symbol)/t
I think r stands for radius and t stands for time. I am not sure how to use this formula.

Yes, that's the same formula I used. The angular velocity \(\omega\) is \(2\pi\) radians (one complete circle) per 95 minutes. The radius of the orbital path is 600 km more than the radius of the Earth.
 
  • #5
MarkFL said:
Yes, that's the same formula I used. The angular velocity \(\omega\) is \(2\pi\) radians (one complete circle) per 95 minutes. The radius of the orbital path is 600 km more than the radius of the Earth.

The only 2 numbers they give are 600km and 95 min.
Here is the problem again.

The reflection telescope is deployed in low Earth orbit(600km) with each orbit lasting about 95 minutes. linear velocity is calculated by the formula

v= radius(radian symbol)/ time.
 
  • #6
dtippitt said:
The only 2 numbers they give are 600km and 95 min.
Here is the problem again.

The reflection telescope is deployed in low Earth orbit(600km) with each orbit lasting about 95 minutes. linear velocity is calculated by the formula

v= radius(radian symbol)/ time.

According to google, the radius of the Earth is about 6,371 km. So, plug that into the formula I posted above...what do you get?
 
  • #7
I think the right answer is 28000 but I don't know how to get that.
 
  • #8
dtippitt said:
I think the right answer is 28000 but I don't know how to get that.

It appears you are to give the answer in km/hr, in which case you would multiply the result you get from the formula I gave by 60. Your not going to get 28000 km/hr exactly, unless you use a value for the radius of the Earth constructed to give you that value for the linear speed. Using the value I cited, I get about 27663 km/hr.
 

1. What is linear velocity in low Earth orbit?

Linear velocity in low Earth orbit refers to the speed at which an object is moving in a straight line as it orbits around the Earth. It is measured in meters per second (m/s) and is affected by factors such as the orbital altitude and the mass of the object.

2. How is linear velocity calculated in low Earth orbit?

To calculate linear velocity in low Earth orbit, you can use the formula: v = √(GM/R), where v is the linear velocity, G is the gravitational constant, M is the mass of the Earth, and R is the distance between the object and the center of the Earth. This formula takes into consideration the gravitational pull of the Earth on the object.

3. Does linear velocity change in low Earth orbit?

Yes, linear velocity can change in low Earth orbit due to the varying distance between the object and the Earth's center. As an object moves closer to the Earth, its linear velocity increases and decreases as it moves further away. This is known as orbital speed or orbital velocity.

4. How does linear velocity affect objects in low Earth orbit?

The linear velocity of an object in low Earth orbit affects its orbital period, which is the time it takes for the object to complete one full orbit around the Earth. A higher linear velocity means a shorter orbital period, while a lower linear velocity means a longer orbital period.

5. Can linear velocity be changed in low Earth orbit?

Yes, linear velocity can be changed in low Earth orbit by altering the object's orbit through the use of thrusters or other propulsion systems. However, this requires precise calculations and adjustments to ensure the object remains in a stable orbit.

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