Discover the Essential Orbital Velocity for Stable Earth Satellite Orbit

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To determine the essential orbital velocity for a stable Earth satellite orbit, altitude is a critical factor, as it influences the satellite's velocity and gravitational force balance. The centripetal force must equal gravitational force for a stable orbit, leading to the formula GMm/r^2 = mv^2/r. A geosynchronous satellite maintains a constant distance from the Earth, resulting in an average velocity of zero with respect to the ground, while orbital velocity is calculated using the circumference of the orbit divided by time. The Moon's orbital velocity is approximately 1 km/s, and even lower velocities can maintain stable orbits, depending on the altitude. Clarification from the instructor on the question's parameters is advisable for accurate calculations.
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I don't know how to answer this question without a given altitude, but since one isn't given then I probably don't need one. Any ideas?

Determine the minimum velocity an Earth satellite must have in order to purse a stable orbit without falling to the ground.
 
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would the period of the satellite be equal to the period of the Earth? If so it would be 24 hours or 86400 seconds (SI units) and how would that be incorporated into the problem?
 
I know the velocity has to be less than 11 km/sec (escape velocity).
 
In order to have a stable orbit, the centripetal force of the satelite must equal to the Gravitational force.
GMm/r^2=mv^2/r
For a satelite orbiting close to the earth, its radius of orbit is approximately equal to the radius of earth.
You can find v then.
 
This question could be phrased better. You're right, you need to know the altitude. You also need to know velocity with respect to what?

A geosynchronous satellite has a velocity of 0 with respect to the ground. The distance between your house and the satellite that feeds you TV is constant, and since average velocity = delta distance/ time, and delta distance is 0, your average velocity is 0.

But orbital velocity is different. You use circumference of the orbit / time to get orbital velocity of a circular orbit. So what is the circumference of the orbit? It depends on the altitude.

The Moon has an orbital velocity of about 1 km/s. Even if you cut it down to 0.5 km/s, it would still be in a stable, but very elliptical orbit.

The Moon is near the Earth's stability radius, which is roughly defined as 1/3 the radius of the Hill Sphere. So if you compute the orbital velocity at apogee of an object whose apogee is around 400,000 km and whose perigee is around 6478 kilometers (radius of Earth + 100 km altitude), that could be the answer. Or, like you said, 0 could be the answer too.

Ask your teacher to clarify the question.
 
Is there another way to answer this question with the information given in the question and any assumptions that can be made from it?
 
I would guess that the assumptions to be made are a low-earth circular orbit. You would not expect the points I brought up to be covered in an introductory physics class.

Just use a little algebra on the formula Harmony gave you to solve for v. Or just look up the orbit velocity formula in your book.
 
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