a) 0.0028 rad/s
b) [tex]23.68 m/s^2[/tex]
c) [tex]0 m/s^2[/tex]
Explanation:
a)
When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by
[tex]\omega = \frac{\theta}{t}[/tex]
where
[tex]\theta[/tex] is the angular displacement
t is the time interval
The angular speed of an object in circular motion can also be written as
[tex]\omega = \frac{v}{r}[/tex] (1)
where
v is the linear speed of the object
r is the radius of the orbit
For the spaceship in this problem we have:
[tex]v=29,960 km/h[/tex] is the linear speed, converted into m/s,
[tex]v=8322 m/s[/tex]
[tex]r=2925 km = 2.925\cdot 10^6 m[/tex] is the radius of the orbit
Subsituting into eq(1), we find the angular speed of the spaceship:
[tex]\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s[/tex]
b)
When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.
The magnitude of the radial acceleration is given by
[tex]a_r=\omega^2 r[/tex]
where
[tex]\omega[/tex] is the angular speed
[tex]r[/tex] is the radius of the orbit
For the spaceship in the problem, we have
[tex]\omega=0.0028 rad/s[/tex] is the angular speed
[tex]r=2925 km = 2.925\cdot 10^6 m[/tex] is the radius of the orbit
Substittuing into the equation above, we find the radial acceleration:
[tex]a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2[/tex]
c)
When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.
The tangential acceleration is given by
[tex]a_t=\frac{\Delta v}{\Delta t}[/tex]
where
[tex]\Delta v[/tex] is the change in the linear speed
[tex]\Delta t[/tex] is the time interval
In this problem, the spaceship is moving with constant linear speed equal to
[tex]v=8322 m/s[/tex]
Therefore, its linear speed is not changing, so the change in linear speed is zero:
[tex]\Delta v=0[/tex]
And therefore, the tangential acceleration is zero as well:
[tex]a_t=\frac{0}{\Delta t}=0 m/s^2[/tex]
