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Do problem 3-15 for a parking orbit at 300 nmi (522 km) and an elliptical orbit with a perigee of 175 nmi (322 km) and an eccentricity of 0.70?

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Final answer:

To calculate the orbital radius for an Earth satellite with a period of 1.00 h, we can use Kepler's third law which relates the period of an orbit to its radius. The result is unreasonable because it is much smaller than the radius of the Earth, indicating that the satellite would be orbiting inside the Earth which is not possible. A 1.00 h orbit is also unreasonable because it violates the laws of physics and the limitations of our current technology.

Step-by-step explanation:

To calculate the orbital radius for an Earth satellite with a period of 1.00 h, we can use Kepler's third law which relates the period of an orbit to its radius. The equation is given as:

T² = k * r³

Where T is the period, r is the radius, and k is a constant.

Rewriting the equation to solve for r, we get:

r = (T²/k)^(1/3)

Using the given period of 1.00 h and solving for the radius, we find that the orbital radius is unreasonable as it is much smaller than the radius of the Earth. This means that the satellite would be orbiting inside the Earth, which is not possible.

The premise of a 1.00 h orbit is also unreasonable because it violates the laws of physics and the limitations of our current technology. Satellites in low Earth orbit have periods of around 90 minutes.

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User Ozoli
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