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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=R√gR+h
or, 6200=6400000∗√106400000+h
or, 9.38∗10−7=√106400000+h
or, h=10655301−6400000
h = 4255301m
Now,
T=2π√(R+h)3g∗R2
or, 2π√(4255301+6400000)310∗6400000
or, 2π∗1718.57=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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