# A satellite was placed in an orbit above the earth's atmosphere, at a distance of 15,000 km from the center of the earth. What is the orbital period (in hours) of the satellite?

The orbital period of this satellite is 5.08 hours (5 hours, 5 minutes, 3 seconds to be precise!) .

Kepler's third law—often referred to as "Newton's derivation"—is required for this problem.

The answer is as follows:

Subtract 3600 (the number of seconds in an hour) to obtain

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The orbital period is

The satellite's velocity is

Consequently,

G is 6.67 * 10^-11 Nm^2 kg^-2.

Thus,

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The orbital period of a satellite in a circular orbit around Earth can be calculated using Kepler's third law of planetary motion. The formula is:

[ T = 2\pi \sqrt{\frac{r^3}{G M}} ]

Where:

- ( T ) is the orbital period (in seconds).
- ( r ) is the distance from the center of the Earth to the satellite (in meters).
- ( G ) is the gravitational constant ((6.674 \times 10^{-11} , \text{m}^3 , \text{kg}^{-1} , \text{s}^{-2})).
- ( M ) is the mass of the Earth ((5.972 \times 10^{24} , \text{kg})).

Converting the distance from kilometers to meters, ( r = 15,000 , \text{km} = 15,000,000 , \text{m} ), we can calculate the orbital period.The orbital period of the satellite is approximately 12.9 hours.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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