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

`a=8.87* 10^(-3)\ km/s^2`

Step-by-step 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=(v^2)/(R)`

a=acceleration

v=speed

Now by putting the values in the above equation we get

`a=(7.71^2)/(6701)\ km/s^2`

`a=0.00887\ km/s^2`

`a=8.87* 10^(-3)\ km/s^2`

Therefore the centripetal acceleration will be

`a=8.87* 10^(-3)\ km/s^2`