1) 0.0011 rad/s
2) 7667 m/s
Explanation:
1)
The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:
[tex]\omega=\frac{\theta}{t}[/tex]
where
[tex]\theta[/tex] is the angular displacement of the object
t is the time elapsed
[tex]\omega[/tex] is the angular velocity
In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is
[tex]\theta=2\pi[/tex] rad
And the time taken is
[tex]t=95 min \cdot 60 =5700 s[/tex]
Therefore, the angular velocity of the telescope is
[tex]\omega=\frac{2\pi}{5700}=0.0011 rad/s[/tex]
2)
For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation
[tex]v=\omega r[/tex]
where
v is the linear velocity
[tex]\omega[/tex] is the angular velocity
r is the radius of the circular orbit
In this problem:
[tex]\omega=0.0011 rad/s[/tex] is the angular velocity of the Hubble telescope
The telescope is at an altitude of
h = 600 km
over the Earth's surface, which has a radius of
R = 6370 km
So the actual radius of the Hubble's orbit is
[tex]r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m[/tex]
Therefore, the linear velocity of the telescope is:
[tex]v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s[/tex]
