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?

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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?

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Answer 1 · 2018-02-05T16:57:42

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"

$T^{2} = (\frac{4 π^{2}}{G M}) r^{3}$ $T^2=((4pi^2)/(GM))r^3$

where G is the gravitational constant $6.67 \times 10^{- 11}$ $6.67xx10^(-11)$ and $M$ $M$ is the mass of the body that is being orbited (Earth in this case).

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

Here is the solution:

$T^{2} = (\frac{4 π^{2}}{(6.67 \times 10^{- 11}) (5.97 \times 10^{24})}) {(1.5 \times 10^{7})}^{3}$ $T^2=((4pi^2)/((6.67xx10^(-11))(5.97xx10^(24))))(1.5xx10^7)^3$

$T^{2} = 3.35 \times 10^{8} s^{2}$ $T^2=3.35xx10^8s^2$

$T = 18292 s$ $T=18 292 s$

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

$T = 5.08$ $T=5.08$ hours

Answer 2 · 2018-02-05T17:14:25

The orbital period is $= 5.077 h$ $=5.077h$

Explanation:

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

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

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

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

Let the gravitational constant be $= G$ $=G$

There are $2$ $2$ forces acting on the satellite, the gravitational force $F_{G}$ $F_G$ anf the centripetal force $F_{C}$ $F_C$

$F_{G} = F_{C}$ $F_G=F_C$

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

The velocity of the satellite is

$v^{2} = \frac{G M}{r}$ $v^2=(GM)/r$

$v = \sqrt{\frac{G M}{r}}$ $v=sqrt((GM)/r)$

The orbital period is $= T s$ $=Ts$

The distance is $d = 2 π r$ $d=2pir$

Therefore,

$v T = 2 π r$ $vT=2pir$

$T = \frac{2 π r}{v} = 2 π r \sqrt{\frac{r}{G M}}$ $T=(2pir)/v=2pirsqrt(r/(GM))$

$T = 2 π \sqrt{\frac{r^{3}}{G M}}$ $T=2pisqrt(r^3/(GM))$

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

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

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

So,

$T = 2 π \sqrt{\frac{{(15 \cdot 10^{6})}^{3}}{6.67 \cdot 10^{- 11} \cdot 5.98 \cdot 10^{24}}}$ $T=2pisqrt((15*10^6)^3/(6.67*10^-11*5.98*10^24))$

$= 18276.9 s$ $=18276.9s$

$= 5.077 h$ $=5.077h$

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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?

2 Answers

Explanation:

Explanation:

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