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An earth satellite moves in a circular orbit with a speed of 6.2km/s. Find the time of one revolution and its centripetal acceleration.

Given,

Orbital velocity vo=6.2km/s=6200m/s
Radius of earth (R)=6400000m
Time Period (T)=?
Centripetal Acceleration (a)=?

We have,
v0=RgR+h
or, 6200=6400000106400000+h
or, 9.38107=106400000+h
or, h=106553016400000
h = 4255301m

Now, 
T=2π(R+h)3gR2
or, 2π(4255301+6400000)3106400000
or, 2π1718.57=10792.65Sec=2.99hrs
∴ Time of revolution = 2.99hrs

Again,
Centripetal Acceleration a=v20h+R=(6200)24255301+6400000=3.6m/s2
∴ Centripetal Acceleration = 3.6m/s2

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