Frequency of the em signal emitted by a rocket I `4xx10^(7)Hz`. If apparent frequency observed on earth is `3.2xx10^(7)Hz`, then velocity with which rocket is moving away is [speed of light = c]

Options

A. 0.5 c
B. 0.7 c
C. 0.9 c
D. 0.2 c

Answer

To find the velocity of the rocket moving away from Earth, we can use the formula for the apparent frequency of a wave when the source is moving away from the observer. The formula is given by:

\[ f' = f \cdot \sqrt{\frac{c - v}{c + v}} \]

Where:
- \( f' \) is the apparent frequency observed on Earth.
- \( f \) is the original frequency emitted by the rocket.
- \( c \) is the speed of light.
- \( v \) is the velocity of the rocket.

**Step 1: Identify the given values**
- Original frequency, \( f = 4 \times 10^7 \, \text{Hz} \)
- Apparent frequency, \( f' = 3.2 \times 10^7 \, \text{Hz} \)
- Speed of light, \( c \) (we will keep it as a variable for now).

**Step 2: Substitute the known values into the formula**
\[ 3.2 \times 10^7 = 4 \times 10^7 \cdot \sqrt{\frac{c - v}{c + v}} \]

**Step 3: Simplify the equation**
Divide both sides by \( 4 \times 10^7 \):
\[ \frac{3.2}{4} = \sqrt{\frac{c - v}{c + v}} \]
This simplifies to:
\[ 0.8 = \sqrt{\frac{c - v}{c + v}} \]

**Step 4: Square both sides to eliminate the square root**
\[ (0.8)^2 = \frac{c - v}{c + v} \]
\[ 0.64 = \frac{c - v}{c + v} \]

**Step 5: Cross-multiply to solve for \( v \)**
\[ 0.64(c + v) = c - v \]
Expanding gives:
\[ 0.64c + 0.64v = c - v \]

**Step 6: Rearrange the equation to isolate \( v \)**
Combine like terms:
\[ 0.64v + v = c - 0.64c \]
\[ (1 + 0.64)v = (1 - 0.64)c \]
\[ 1.64v = 0.36c \]

**Step 7: Solve for \( v \)**
\[ v = \frac{0.36}{1.64}c \]
Calculating the fraction:
\[ v \approx 0.2195c \]

**Final Answer:**
The velocity of the rocket moving away from Earth is approximately \( 0.22c \).