Question

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

Answer 1

Considering that 
$1 day = (24 h/d) ( 60 min/h)(60 s/min)=86400 s$,
the frequency of revolution of the international space spation around Earth is
$f= \frac{15.65 rev}{86400 s}=1.8 \cdot 10^{-4}Hz$
and so its orbital period is
$T= \frac{1}{f}= \frac{1}{1.8 \cdot 10^{-4} Hz}=5521 s$

Now we can use Kepler's third law to find the radius of the orbit of the space station, r. Knowning that the Earth mass is $M=5.97 \cdot 10^{24}kg$, Kepler's third law can be written as
$\frac{r^3}{T^2}= \frac{GM}{4 \pi^2}$
where G is the gravitational constant. Re-arranging the formula and using the value of T we found previously, we find the orbital radius:
$r= \sqrt[3]{ \frac{GMT^2}{4 \pi^2} }= \sqrt[3]{ \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)(5521s)^2}{4 \pi^2} } =6.75 \cdot 10^6 m$
which corresponds to r=6750 km.

This is the radius of the orbit measured from the center of Earth. To find the distance of the space station from the Earth's surface, we must subtract the Earth's radius, R=6370 km, and we find:
$h=r-R=6750 km-6370 km=380 km$

So, the space station is approximately 380 km above Earth's surface.

Answer 2

This satellite is about 3.79 × 10⁵ m above the surface of the earth

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Further explanation

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

$\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }$

F = Gravitational Force ( Newton )

G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )

m = Object's Mass ( kg )

R = Distance Between Objects ( m )

Let us now tackle the problem !

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Given:

orbital period of asteroid = T = (1 × 24 × 3600)/(15.65) ≈ 5.521 × 10³ s

mass of Earth = M = 5.972 × 10²⁴ kg

radius of Earth = R_earth = 6.371 × 10⁶ m

Asked:

height position of space station = h = ?

Solution:

Firstly, we will calculate the orbital radius of the space station from centre of the Earth:

$\Sigma F = ma$

$G \frac{ M m} { R^2 } = m \omega^2 R$

$G \frac{ M } { R^2 } = \omega^2 R$

$G \frac{ M } { R^3 } = \omega^2$

$G \frac{ M } { R^3 } = (2\pi \div T)^2$

$R = \sqrt[3] { GM (\frac{T}{2\pi})^2 }$

$R = \sqrt[3] { 6.67 \times 10^{-11} \times 5.972 \times 10^{24} \times (\frac{5.521 \times 10^3}{2\pi})^2 }$

$\boxed{R \approx 6.75 \times 10^{6} \texttt{ m}}$

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Next, we could calculate the height position of the space station:

$h = R - R_{earth}$

$h = 6.75 \times 10^6 - 6.371 \times 10^6$

$\boxed{h \approx 3.79 \times 10^5 \texttt{ m }}$

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Learn more

• Impacts of Gravity : https://brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
• The Acceleration Due To Gravity : https://brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Gravitational Fields