Two galaxies on opposite ends of the universe are moving away from the Earth. Each has a velocity of 200,000 km/s relative to the Earth. How fast would an observer in one of those galaxies see the other galaxy moving away? (1 point)

Responses

200,000 km/s
200,000 km/s

between 300,000 and 400,000 km/s
between 300,000 and 400,000 km/s

between 200,000 and 300,000 km/s
between 200,000 and 300,000 km/s

400,000 km/s

1 answer

To determine how fast an observer in one galaxy would see the other galaxy moving away, we need to use the relativistic velocity addition formula, since the speeds involved are a significant fraction of the speed of light (which is approximately 300,000 km/s).

The relativistic formula for adding velocities \( u \) and \( v \) is given by:

\[ w = \frac{u + v}{1 + \frac{uv}{c^2}} \]

Where:

  • \( w \) is the relative velocity of one object as seen from the other,
  • \( u \) is the velocity of the first object relative to a third reference point (in this case, Earth),
  • \( v \) is the velocity of the second object relative to the same reference point (Earth),
  • \( c \) is the speed of light.

In this case, both galaxies are moving away from Earth at a velocity of \( 200,000 \) km/s.

Let's use \( u = 200,000 \) km/s and \( v = 200,000 \) km/s.

Now substituting these values in the formula, we first need \( c^2 \):

\[ c^2 = (300,000)^2 \]

Now we can put the values into the relativistic addition formula:

\[ w = \frac{200,000 + 200,000}{1 + \frac{(200,000)(200,000)}{(300,000)^2}} \]

Calculating the denominator:

\[ \frac{(200,000)(200,000)}{(300,000)^2} = \frac{40,000,000,000}{90,000,000,000} = \frac{4}{9} \]

Now substitute this back into the equation for \( w \):

\[ w = \frac{400,000}{1 + \frac{4}{9}} = \frac{400,000}{\frac{13}{9}} = 400,000 \times \frac{9}{13} \approx 276,923 \text{ km/s} \]

Thus, the velocity of one galaxy as seen from the other galaxy is approximately \( 276,923 \) km/s.

So, the answer would be between 200,000 and 300,000 km/s.