Question

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

Answer 1

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**.