Question

Near the center of our galaxy, hydrogen gas is moving directly away from us in its orbit about a black hole. We receive 1900 nm electromagnetic radiation and know that it was 1875 nm when emitted by the hydrogen gas. What is the speed of the gas?

Answer 1

Speed of gas = 3.973 x 10^(6) m/s

Step-by-step explanation:

From the relativistic doppler effect, the observed frequency is given as;

λobs = λs√[(1 + (v/c))/(1-(v/c))]

Where;

λobs is the wavelength received by the observer

λs is the wavelength of the wave emitted from the source

v is the relative velocity between the observer and the source

c is the speed of light which is 3 x 10^(8) m/s

λobs = 1900 nm

λs = 1875 nm

Let's make the relative velocity(v) the subject of the equation.

λobs = λs√[(1 + (v/c))/(1-(v/c))]

(λobs/λs)²= [(1 + (v/c))/(1-(v/c))]

Let's call (λobs/λs) = p

Thus;

p² = [(1 + (v/c))/(1-(v/c))]

p²[(c-v)/c] = (c+v)/c

Multiply both sides by c to obtain ;

p²(c-v) = c + v

p²c - p²v = c + v

p²c - c = v(p² + 1)

v = c(p²-1)/(p²+1)

Now since p = (λobs/λs)

p = 1900/1875 = 1.0133

So, v = 3x10^(8) [1.0133²-1]/[1.0133²+1]

= 3.973 x 10^(6) m/s

Answer 2

The speed of the hydrogen gas is 4000000m/s.

Step-by-step explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest for this case is 1875 nm (
`\lambda_(0) = 1875nm`)

`Redshift: \lambda_(measured)  >  \lambda_(0)`

`Blueshift: \lambda_(measured)  <  \lambda_(0)`

Since,
`\lambda_(measured)` (1900nm) is greater than
`\lambda_(0)` (1875 nm), it can be confirmed that the hydrogen gas is moving away from the observer as the statement says.

Due to that shift, the speed of the hydrogen gas can be determined by means of the Doppler velocity.

`v = c(\Delta \lambda)/(\lambda_(0))` (1)

Where
`\Delta \lambda` is the wavelength shift,
`\lambda_(0)` is the wavelength at rest, v is the speed of the source and c is the speed of light.

`v = c((\lambda_(measured)- \lambda_(0))/(\lambda_(0)))` (2)

`v = (3x10^(8)m/s)((1900nm- 1875nm)/(1875nm))`

`v = 4000000m/s`

Hence, the speed of the hydrogen gas is 4000000m/s.