# 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?

Feb 5, 2018

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

#### Explanation:

We need Kepler's third law for this problem, but in the form that is often referred to as "Newton's derivation"

${T}^{2} = \left(\frac{4 {\pi}^{2}}{G M}\right) {r}^{3}$

where G is the gravitational constant $6.67 \times {10}^{- 11}$ and $M$ is the mass of the body that is being orbited (Earth in this case).

Also, care must be taken to use base units for $T , r \mathmr{and} M$, as the value stated for $G$ includes base units only.

Here is the solution:

${T}^{2} = \left(\frac{4 {\pi}^{2}}{\left(6.67 \times {10}^{- 11}\right) \left(5.97 \times {10}^{24}\right)}\right) {\left(1.5 \times {10}^{7}\right)}^{3}$

${T}^{2} = 3.35 \times {10}^{8} {s}^{2}$

$T = 18 292 s$

Divide by 3600 (seconds in each hour) and you get

$T = 5.08$ hours

Feb 5, 2018

The orbital period is $= 5.077 h$

#### Explanation:

Let the mass of the earth $= M k g$

Let the mass of the satellite be $= m k g$

Let the radius of the earth be $= R m$

Let the radius of the satellite's orbit be $= r m$

Let the gravitational constant be $= G$

There are $2$ forces acting on the satellite, the gravitational force ${F}_{G}$ anf the centripetal force ${F}_{C}$

${F}_{G} = {F}_{C}$

$G M \frac{m}{r} ^ 2 = m {v}^{2} / r$

The velocity of the satellite is

${v}^{2} = \frac{G M}{r}$

$v = \sqrt{\frac{G M}{r}}$

The orbital period is $= T s$

The distance is $d = 2 \pi r$

Therefore,

$v T = 2 \pi r$

$T = \frac{2 \pi r}{v} = 2 \pi r \sqrt{\frac{r}{G M}}$

$T = 2 \pi \sqrt{{r}^{3} / \left(G M\right)}$

$r = 15000 k m = 15 \cdot {10}^{6} m$

$G = 6.67 \cdot {10}^{-} 11 N {m}^{2} k {g}^{-} 2$

$M = {5.9810}^{24} k g$

So,

$T = 2 \pi \sqrt{{\left(15 \cdot {10}^{6}\right)}^{3} / \left(6.67 \cdot {10}^{-} 11 \cdot 5.98 \cdot {10}^{24}\right)}$

$= 18276.9 s$

$= 5.077 h$