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Q2.31:
The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us ?

Solution :

Step 1: Identifying Given Variables and Physical Constants

To calculate the distance $d$ of the quasar, we utilize the relationship between speed, time, and distance: $d = v \times t$.

  • Speed of light ($c$): $3.0 \times 10^8 \text{ m/s}$
  • Time taken ($t$): $3.0 \text{ billion years} = 3.0 \times 10^9 \text{ years}$
  • Conversion factor (1 year to seconds): $1 \text{ year} \approx 365.25 \text{ days} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} \approx 3.15576 \times 10^7 \text{ seconds}$

Step 2: Performing Unit Conversions

First, convert the time $t$ from years into seconds:

$t = (3.0 \times 10^9 \text{ years}) \times (3.15576 \times 10^7 \text{ s/year})$ $t = 9.467 \times 10^{16} \text{ s}$ [By dimensional consistency]

Step 3: Calculating Distance in Meters

Apply the formula $d = c \times t$:

$d = (3.0 \times 10^8 \text{ m/s}) \times (9.467 \times 10^{16} \text{ s})$ $d = 2.8401 \times 10^{25} \text{ meters}$

Step 4: Converting Meters to Kilometers

Since $1 \text{ km} = 10^3 \text{ m}$, we adjust the exponent:

$d = \frac{2.8401 \times 10^{25}}{10^3} \text{ km}$ $d = 2.84 \times 10^{22} \text{ km}$ [Applying significant figure rules for 3.0]

Visual Representation of Light Propagation

Quasar Earth Distance $d = 2.84 \times 10^{22}$ km Light travel time $t = 3.0 \times 10^9$ years
Variable Value
Speed of Light ($c$) $3.0 \times 10^8 \text{ m/s}$
Time ($t$) $9.47 \times 10^{16} \text{ s}$
Distance ($d$) $2.84 \times 10^{22} \text{ km}$

Final Solution: The distance of the quasar from Earth is approximately $2.84 \times 10^{22} \text{ km}$.


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