Question

A hydrogen line in a star's spectrum has a frequency of 6.17*10^14 Hz when stationary. In Rigel's Spectrum, it is shifted downward by 4.26*10^10Hz. What is Rigel's velocity relative to us? m/s

Answer 1

Rigel's velocity relative to us is approximately -20713.13 m/s. The negative sign indicates that Rigel is moving away from us.

How can you solve Rigel's velocity relative to us?

The observed frequency in Rigel's spectrum is shifted downward by 4.26 * 10¹⁰ Hz compared to the stationary frequency. This difference is called the redshift.

The Doppler shift formula relates the observed frequency (f obs), the stationary frequency (f 0), and the velocity (v) of the source:

f obs = f 0 * (1 - v/c)

where c is the speed of light (approximately 3 * 10⁸ m/s).

We can rearrange the formula to solve for the velocity:

v = c * (1 - f obs / f 0)

v = 3 * 10⁸ m/s * (1 - (6.17 * 10¹⁴ Hz - 4.26 * 10¹⁰ Hz) / 6.17 * 10¹⁴ Hz)

v = 3 * 10⁸ m/s * (1 - 5.744 * 10¹⁰ Hz / 6.17 * 10¹⁴ Hz)

v ≈ -20713.13 m/s

Therefore, Rigel's velocity relative to us is approximately -20713.13 m/s. The negative sign indicates that Rigel is moving away from us.

Answer 2

The correct answer will be "2.97 × 10⁸ m/s".

Explanation:

The give values are:

Observed frequency,
`F=4.26* 10^(10) \ Hz`

Original frequency,
`F_(0)=6.17* 10^(14) \ Hz`

Let the velocity of Rigel be = V m/s

As we know,

⇒
`F_(0)=F\sqrt{(C+V)/(C-V)}\\`

On putting the values, we get

⇒
`(6.17* 10^(14) )/(4.26* 10^(10) ) =\sqrt{(C+V)/(C-V) }`

⇒
`(C+V)/(C-V)=(14483.56808)^2`

⇒
`(C+V)/(C-V)=209773744.2`

⇒
`C+V=209773744.2 \ C-209773744.2 \ V`

⇒
`C-209773744.2 \ C=209773744.2 \ V-V`

⇒
`V=(209773743.2)/(209773745.2) \ C`

⇒
`V=0.99 \ C`

So that the Rigel's velocity will be "0.99 C".

Now,

⇒
`V=0.99* 3* 10^8`

⇒
`=2.97* 10^8 \ m/s`