Question

The International Space Station orbits at an average height of 350 km above sea level.???? = 6.67× 10−11 ????m2 ????????2 ⁄ ????????????????????ℎ = 5.97 × 1024 ???????? ????????????????????ℎ = 6.37× 103 ????m(a) Determine the orbital speed of this space station.(b) The period of the space station.Draw it out for full credit

Answer 1

Answer:

i agree with the other persons answer

Explanation:

Answer 2

Answer:

(a). The orbital speed of this space station is 7906.42 m/s.

(b).  The period of the space station is 5062.2 sec.

Explanation:

Given that,

Gravitational constant $G=6.67\times10^{-11}\ Nm^2/kg^2$

Mass of earth $M_{e}=5.97\times10^{24}\ kg$

Radius of earth $R_{e}=6.37\times10^{3}\ km$

(a). We need to calculate the orbital speed of this space station

Using formula of orbital speed

$v=\sqrt{\dfrac{GM}{r}}$

$v=\sqrt{\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{6.37\times10^{6}}}$

$v=7906.42\ m/s$

(b). We need to calculate the period of the space station

Using formula of period

$T=\dfrac{2\pi r}{v}$

Put the value into the formula

$T=\dfrac{2\pi\times6.37\times10^{6}}{7906.42}$

$T=5062.2\ sec$

Hence, (a). The orbital speed of this space station is 7906.42 m/s.

(b).  The period of the space station is 5062.2 sec.