What is the Period of an Earth Satellite Orbiting 400km Above the Surface?

In summary: To clarify, the formula for the perimeter of a circle is 2*pi*r, not 2*pi*r^2. In summary, the task is to find the speed of an Earth satellite at 400km above the Earth's surface and the period of its orbit. The formula used for the first part is v=[(Gm)/r]^(1/2) and the answer is approximately 7.7*10^3 km/s. For the second part, there was confusion with the formula for the perimeter of a circle, which is 2*pi*r, not 2*pi*r^2. The correct answer for the period is approximately 5.6*10^3 s.
  • #1
Torquescrew
17
0

Homework Statement


Find the speed of an Earth satellite whose orbit is 400km above the Earth's surface. What is the period of the orbit.

Homework Equations


The first part was pretty easy. I used v=[(Gm)/r]^(1/2) to get the answer.


The Attempt at a Solution


For the first part, I dropped the values into the equation to get
(6.673*10^-11)(5.97*10^24)
(6.37*10^6 + 400000)^2

I came up with about 7671 m/s (7.7*10^3 km/s) <--tangential velocity, right?


But for the second part, I couldn't find any useful details in my notes. I figured I'd get the circumference and divide by my velocity, but that didn't work.
2(pi)(6.37*10^6 + 400000)^2
7671

I wound up with some huge number.

I also tried Kepler's 3rd law (t^2=Kr^3), which doesn't help much, because I don't have a constant.
The thing is, I already know what the answer is supposed to be. I just don't know how to get it.

The answer is supposed to be approx 5.6*10^3 s (about 93 minutes).

I know I'm close. I tried to reverse engineer the problem by solving for x
5.6*10^3=[2(pi)x]/7671.02

I wound up with x = 6.836*10^6, which is remarkably close to 6.37*10^6 + 400000 but that seems inconsistent with the whole two-pi-arr-squared thing.

So I guess my real question is, if the radius is right, and if my velocity is right, why should I use 2(pi)r instead of 2(pi)r^2? What did I miss?
 
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  • #2
You are on the right path.
You have a mistake in the formula for the perimeter of the circle.
It's 2*pi*r
Not r^2. (r^2 is for area, r^3 for volume)
 
  • #3
Good catch, nasu. And thanks for pointing that out.
 

Related to What is the Period of an Earth Satellite Orbiting 400km Above the Surface?

What is orbital period?

Orbital period refers to the amount of time it takes for an object to complete one full orbit around another object.

What factors affect orbital period?

The primary factors that affect orbital period include the mass of the objects involved, the distance between them, and the gravitational force between them.

Why is difficulty with orbital period important to study?

Understanding the orbital period of objects is crucial in predicting and tracking their movements, which is essential for space exploration and satellite communication.

How is orbital period calculated?

Orbital period can be calculated using Kepler's third law, which states that the square of the orbital period is directly proportional to the cube of the semi-major axis of the orbit.

Can orbital period change over time?

Yes, orbital period can change over time due to various factors such as gravitational interactions with other objects, tidal forces, and atmospheric drag.

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