Question

(I) A space shuttle releases a satellite into a circular orbit 780 km above the Earth. How fast must the shuttle be moving (relative to Earth’s center) when the release occurs?

Answer 1

Answer:

Explanation:

Given

shuttle is at distance of 780 km above earth surface

total distance $r=r_e+d_s$

radius of earth $r_e=6371 km$

$d_s=distance\ of\ shuttle\ from\ earth\ surface$

$r=6371+780=7151 km$

acceleration due to gravity

$g=\frac{Gm_e}{r^2}$

$g=\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{(7151\times 10^3)^2}$

$g=7.73 m/s^2$

where $m_e=mass\ of\ earth$

Net force on shuttle is

$F=ma=mg$

Centripetal acceleration is given by

$F=\frac{mv^2}{r}$

Force will Provide centripetal acceleration

$mg=\frac{mv^2}{r}$

$v=\sqrt{gr}$

$v=\sqrt{7.73\times 7151\times 10^3}$

$v=7.44\times 10^3 m/s$