A satellite circles the earth in an orbit whose radius is 2.01 times the earth's radius. The earth's mass is 5.98 x 1024 kg, and its radius is 6.38 x 106 m. What is the period of the satellite?

To find the period of the satellite, we need to use the formula for the period of a satellite in circular orbit:

T = 2π * √(r³ / GM)

where T is the period, π is a constant approximately equal to 3.14159, r is the radius of the orbit, G is the gravitational constant, and M is the mass of the earth.

Let's plug in the given values:

r = 2.01 * 6.38 x 10^6 m = 12.8038 x 10^6 m
M = 5.98 x 10^24 kg
G = 6.67430 x 10^-11 m³/(kg s²)

Now we can calculate the period of the satellite:

T = 2π * √((12.8038 x 10^6)³ / (6.67430 x 10^-11 * 5.98 x 10^24))

To simplify the calculation, let's first compute the value inside the square root:

(12.8038 x 10^6)³ = 2.785 x 10^20

Now we can substitute the values and calculate:

T = 2π * √(2.785 x 10^20 / (6.67430 x 10^-11 * 5.98 x 10^24))

Dividing the numerator and denominator by 5.98 x 10^24, we get:

T = 2π * √(2.785 x 10^20 / 6.67430 x 10^-11)

T = 2π * √(2.785 x 10^20 * 1.49915 x 10^10)

T = 2π * √(4.178432275 x 10^30)

T = 2π * 2.044936673 x 10^15

Using a calculator to evaluate this expression, we get:

T = 2.191 x 10^16 s

Therefore, the period of the satellite is approximately 2.191 x 10^16 seconds.

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