Answer to Question #84643 in Physics for felix

Question

a spy satellite is in circular orbit around earth. it makes one revolution in 6.00 hr, how high above the earth's surface is the satellite?

Answer 1

Let's first find the distance of the satellite from the center of the Earth from the Kepler's Third Law:

$$\frac{T^2}{R^3} = \frac{4\pi^2}{GM_E},$$

here,

$$T$$

is the period that the satellite needs to make one revolution,

$$R$$

is the distance of the satellite from the center of the Earth,

$$G$$

is the universal gravitational constant and

$$M_E$$

is the mass of the Earth.

Then, from this equation we can find the distance

of the satellite from the center of the Earth:

$$R = \sqrt[3]{\frac{GMT^2}{4\pi^2}}.$$

Let's substitute the numbers:

$$R = \sqrt[3]{\frac{6.67 \cdot 10^{-11} \frac{Nm^2}{kg^2} \cdot 5.98 \cdot 10^{24} kg \cdot (6.0 hr \cdot \frac{3600 s}{1 hr})^2}{4\pi^2}} = 1.67 \cdot 10^7 m.$$

We can find the height of the satellite above the Earth's surface from the formula:

$$R = R_{Earth} + h,$$

here,

$$R_{Earth}$$

is the radius of the Earth,

$$h$$

is the height of the satellite above the Earth's surface.

Finally, we get:

$$h = R - R_{Earth} = 1.67 \cdot 10^7 m - 6.37 \cdot 10^6 m = 1.03 \cdot 10^7 m.$$

Answer:

$$h = 1.03 \cdot 10^7 m$$

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