Question

The international space station makes 15.65 revolutions per day in its orbit around the earth. assuming a circular orbit, how high is this satellite above the surface of the earth?

Answer 1

The International space station is located at the height of $\boxed{369\,{\text{km}}}$  above the surface of the Earth.

Further Explanation:

The height of the international space station above the surface of the earth is given by the Kepler’s Law of planetary motion. According to this law, the square of time period of the satellite is directly proportional to the cube of the radius of the circular path of the satellite.

Given:

The speed of revolutions made by the International space station is $15.65\,{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{{\text{day}}}}}\right.\kern-\nulldelimiterspace}{{\text{day}}}}$ .

Concept:

The angular speed of rotation of the International space station is:

$\begin{aligned}\omega&=2\pi\times\frac{{15.65}}{{24\times60\times 60}}\,{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{{\text{sec}}}}}\right.\kern-\nulldelimiterspace}{{\text{sec}}}}\\&=2\pi\times1.811\times{10^{ - 4}}\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{{\text{sec}}}}}\right.\kern-\nulldelimiterspace}{{\text{sec}}}}\\\end{aligned}$

The time period of rotation of the International space station is:

$T=\frac{{2\pi }}{\omega }$

Substitute $\omega$  in above expression.

$\begin{aligned}T&=\frac{{2\pi}}{{2\pi\times1.811\times{{10}^{ - 4}}\,}}\\&=5520.7\,{\text{s}}\\&\approx{\text{5521}}\,{\text{s}}\\\end{aligned}$

The expression for the Kepler’s law is:

$\begin{aligned}{T^2}&=\left({\frac{{4{\pi ^2}}}{{GM}}}\right){R^3}\\R&={\left({\frac{{GM{T^2}}}{{4{\pi^2}}}}\right)^{\frac{1}{3}}}\\\end{aligned}$

Here, $G$  is the gravitational constant, $M$  is the mass of the Earth.

Substitute the values in above expression.

$\begin{aligned}R&={\left({\frac{{\left( {6.67\times{{10}^{ - 11}}}\right)\times\left({5.98\times{{10}^{24}}}\right)\times{{\left( {5521}\right)}^2}}}{{4\times{{\left({3.14}\right)}^2}}}}\right)^{\frac{1}{3}}}\\&={\left({\frac{{1.21\times{{10}^{22}}}}{{39.48}}}\right)^{\frac{1}{3}}}\\&=6.74\times{10^6}\,{\text{m}}\\\end{aligned}$

The radius of the Earth is $6.371\times{10^6}\,{\text{m}}$ .

The height of space station above the surface of Earth is given below:

$\begin{aligned}h&=\left({6.74\times{{10}^6}}\right)-\left({6.371\times{{10}^6}}\right)\\&=3.69\times{10^5}\,{\text{m}}\\&=3{\text{69}}\,{\text{km}}\\\end{aligned}$

Thus, the International space station is located at the height of  $\boxed{369\,{\text{km}}}$ above the surface of the Earth.

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Gravitation

Keywords:

International space station, 15.65 revolutions, circular orbit, around the earth, high, satellite, above the surface, 5521 s, 369 km.

Answer 2

373.2 km The formula for velocity at any point within an orbit is v = sqrt(mu(2/r - 1/a)) where v = velocity mu = standard gravitational parameter (GM) r = radius satellite currently at a = semi-major axis Since the orbit is assumed to be circular, the equation is simplified to v = sqrt(mu/r) The value of mu for earth is 3.986004419 Ă— 10^14 m^3/s^2 Now we need to figure out how many seconds one orbit of the space station takes. So 86400 / 15.65 = 5520.767 seconds And the distance the space station travels is 2 pi r, and since velocity is distance divided by time, we get the following as the station's velocity 2 pi r / 5520.767 Finally, combining all that gets us the following equality v = 2 pi r / 5520.767 v = sqrt(mu/r) mu = 3.986004419 Ă— 10^14 m^3/s^2 2 pi r / 5520.767 s = sqrt(3.986004419 * 10^14 m^3/s^2 / r) Square both sides 1.29527 * 10^-6 r^2 s^2 = 3.986004419 * 10^14 m^3/s^2 / r Multiply both sides by r 1.29527 * 10^-6 r^3 s^2 = 3.986004419 * 10^14 m^3/s^2 Divide both sides by 1.29527 * 10^-6 s^2 r^3 = 3.0773498781296 * 10^20 m^3 Take the cube root of both sides r = 6751375.945 m Since we actually want how far from the surface of the earth the space station is, we now subtract the radius of the earth from the radius of the orbit. For this problem, I'll be using the equatorial radius. So 6751375.945 m - 6378137.0 m = 373238.945 m Converting to kilometers and rounding to 4 significant figures gives 373.2 km