Question

The International Space Station (ISS) flies on a circular orbit with a speed of 7.71 km/s at a height of 330.0 km above the surface of the Earth. What is the centripetal acceleration of the station

Answer 1

Answer:

$a=8.87\times 10^{-3}\ km/s^2$

Explanation:

Given that

Speed ,v= 7.71 km/s

height ,h = 330 km

We know that ,radius of the earth ,r= 6371 km

R= h + r

R= 330 + 6371 = 6701 km

We know that centripetal acceleration is given as

$a=\dfrac{v^2}{R}$

a=acceleration

v=speed

Now by putting the values in the above equation we get

$a=\dfrac{7.71^2}{6701}\ km/s^2$

$a=0.00887\ km/s^2$

$a=8.87\times 10^{-3}\ km/s^2$

Therefore the centripetal acceleration will be

$a=8.87\times 10^{-3}\ km/s^2$