Respuesta :
380 km
The period of an orbit is expressed by the equation
p = sqrt((4*pi^2/u)*a^3)
where
p = period of orbit
u = standard gravitational parameter (product of G and M. For Earth, u is 3.986004418Ă—10^14 m^3/s^2. u is much better to use than either G or M since it's known to a higher precision).
a = semi-major axis of orbit.
Solving for a, gives
p = sqrt((4*pi^2/u)*a^3)
p^2 = (4*pi^2/u)*a^3
up^2/(4pi^2) = a^3
(up^2/(4pi^2))^(1/3) = a
((p^2)u/(4pi^2))^(1/3) = a
Substituting constants
((p^2)3.986004418Ă—10^14 m^3/s^2/(4*3.141592654^2))^(1/3) = a
((p^2)1.009666714x10^13 m^3/s^2)^(1/3) = a
There are 86400 seconds in a day, so the orbital period of the space station is 86400/15.65 = 5520.766773 seconds.
Substituting the value for the period and solving gives:
((p^2)1.009666714x10^13 m^3/s^2)^(1/3) = a
(((5520.766773 s)^2)1.009666714x10^13 m^3/s^2)^(1/3) = a
((30478865.76 s^2)1.009666714x10^13 m^3/s^2)^(1/3) = a
(30478865.76 *1.009666714x10^13 m^3)^(1/3) = a
(3.077349624x10^20 m^3)^(1/3) = a
6751375.759 m = a
6751.375759 km = a
So the semi-major axis is about 6751 kilometers. Now we need to subtract the radius of the earth to get the altitude of the space station. So
6751.375759 km - 6371 km = 380.375 km
Rounding to 4 significant figures gives 380 km.
The International space station will be at altitude of [tex]\boxed{369\,{\text{km}}}[/tex] from the Earth’s surface.
Further Explanation:
The altitude of the ISS from the Earth is given by the Kepler’s Law of planetary motion. According to this law, the square of the time taken by the satellite to complete one revolution is directly proportional to the cube of the orbital radius.
Given:
The time taken by International space station to complete revolutions is [tex]15.65\,{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{{\text{day}}}}}\right.\kern-\nulldelimiterspace}{{\text{day}}}}[/tex] .
Concept:
In one day, the number of seconds is:
[tex]\begin{aligned}1\,{\text{day}}&=\left({\frac{{24\,{\text{hr}}}}{{1\,{\text{day}}}}}\right)\times \left({\frac{{60\,\min }}{{1\,{\text{hr}}}}}\right)\times\left({\frac{{60\,\sec }}{{1\,\min }}}\right)\\&=24\times3600\,{{{\text{sec}}}\mathord{\left/{\vphantom{{{\text{sec}}}{{\text{day}}}}}\right.\kern-\nulldelimiterspace}{{\text{day}}}}\\&=86400\,\sec\\\end{aligned}[/tex]
The speed by which the ISS completes its revolution is termed as its angular speed. It is given as:
[tex]\begin{aligned}\omega&=2\pi\times\frac{{15.65}}{{86400}}\,{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{{\text{sec}}}}}\right.\kern-\nulldelimiterspace}{{\text{sec}}}}\\&=2\pi\times 1.811\times {10^{ - 4}}\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{{\text{sec}}}}}\right.\kern-\nulldelimiterspace}{{\text{sec}}}}\\\end{aligned}[/tex]
The time taken by the International space station to complete one revolution is:
[tex]T=\frac{{2\pi }}{\omega }[/tex]
Substitute [tex]\omega[/tex] in above expression.
[tex]\begin{aligned}T&=\frac{{2\pi }}{{2\pi\times1.811\times {{10}^{ - 4}}\,}}\\&=5520.7\,{\text{s}}\\&\approx{\text{5521}}\,{\text{s}}\\\end{aligned}[/tex]
The Kepler’s law of planetary motion can be expressed as:
[tex]\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}[/tex]
Here, [tex]G[/tex] is the gravitational constant, [tex]M[/tex] is the mass of the Earth.
Substitute the values in above expression.
[tex]\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}[/tex]
The radius of the Earth is [tex]6.371\times{10^6}\,{\text{m}}[/tex] .
The altitude of the international space station above the Earth is:
[tex]\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}[/tex]
Thus, the International space station will be at altitude of [tex]\boxed{369\,{\text{km}}}[/tex] from the Earth’s surface.
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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.
