# Suppose a planet orbits its star in a circular orbit (uniform circular motion) of radius 1.62 x...

## Question:

Suppose a planet orbits its star in a circular orbit (uniform circular motion) of radius 1.62 {eq}\times {/eq} 10{eq}^{11} {/eq} m. The orbital period of the planet around its star is 37.0 years. Determine the following quantities for this orbital motion:

a. angular acceleration

b. tangential acceleration

c. radial acceleration

d. angular velocity

e. tangential velocity

## Stable Orbit

For a mass m to be in a stable orbit about a larger mass M, the centripetal acceleration must be equal to the gravitational acceleration exerted by the larger mass.

Centripetal acceleration is given by:

{eq}a_c=\dfrac{v_t^2}{r} {/eq}

Where {eq}v_t {/eq} is the tangential velocity, and r is the center-to-center distance.

Gravitational acceleration is given by:

{eq}g=\frac{GM}{r^2} {/eq}

Where the gravitational constant is:

{eq}G=6.674\times 10^{-11} \frac{m^3}{kg \cdot s^2} {/eq}

## Answer and Explanation:

Uniform circular motion mean the angular velocity is not changing, this means there is no angular acceleration.

- a. {eq}\boxed{\alpha=0} {/eq}

If the anguular velocity is not changing, the tangential velocity is also constant, meaning no tangential acceleration.

- b. {eq}\boxed{a_t=0} {/eq}

Uniform circular motion requires a continuous acceleration (centripetal) in the radial direction. However, to calculate that, we'll need the tangential velocity, which requires the angular velocity. So we'll come back to part c.

Angular velocity is related to period by:

{eq}\omega=2\pi T {/eq}

The given period in seconds is:

{eq}T=37.0 years 1.167 \times 10^{9} s {/eq}

Which gives an angular velocity of:

{eq}\omega=\frac{2\pi}{T}=5.384 \times 10^{-9} \frac{rad}{s} {/eq}

- d. {eq}\boxed{\omega= 5.384 \times 10^{-9}\frac{rad}{s}} {/eq}

This is related to tangential velocity by:

{eq}\omega=\frac{v_t}{r} {/eq}

So the tangential velocity is:

{eq}v_t=\omega r = 5.384 \times 10^{-9}\frac{rad}{s} (1.62 \times 10^{11} m)=872 \frac{m}{s} {/eq}

- e. {eq}\boxed{v_t=872 \frac{m}{s}} {/eq}

Centripetal acceleration is given by:

{eq}a_c=\frac{v_t^2}{r}=\frac{872^2}{(1.62 \times 10^{11} )}=4.7\times 10^{-6} \frac{m}{s^2} {/eq}

- c. {eq}\boxed{a_c=4.7\times 10^{-6} \frac{m}{s^2}} {/eq}