Question

Question: (a) How long does it take a radio signal to travel $\mathbf{150}\mathbf{}\mathbf{km}$from a transmitter to a receiving antenna? (b) We see a full Moon by reflected sunlight. How much earlier did the light that enters our eye leave the Sun? The Earth–Moon and Earth–Sun distances are $\mathbf{3}\mathbf{.}\mathbf{8}\mathbf{x}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{km}$ and , respectively. (c) What is the round-trip travel time for light between Earth and a spaceship orbiting Saturn, $\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{x}{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathbf{km}$ distant? (d) The Crab nebula, which is about $\mathbf{6500}$light years (ly) distant, is thought to be the result of a supernova explosion recorded by Chinese astronomers in A.D. 1054. In approximately what year did the explosion actually occur? (When we look into the night sky, we are effectively looking back in time.)

Short answer

1. Time taken by radio signal to travel $150\mathrm{km}$is $5.0\mathrm{x}{10}^{-4}\mathrm{s}$.
2. Time taken by the sunlight to reflect from moon and enter our eye is $8.4\min$.
3. Round-trip travel time for light between Earth and a spaceship orbiting Saturn is $2.4\mathrm{hours}$.
4. The year in which the supernova explosion which resulted in Crab nebula occur is $5446\mathrm{BC}$.

Step-by-step solution

Given data:

Distance traveled by radio signal is$150\mathrm{km}$.

Earth-Moon distance is$3.8x{10}^5\mathrm{km}$.

Earth-Sun distance is$1.5x{10}^8\mathrm{km}$.

Distance between Earth and spaceship is $1.3x{10}^9\mathrm{km}$.

Distance of Crab nebula is $6500$light years.

 Step 2: Understanding the concept:

By using the relation between time, distance, and velocity, you can find the required answers.

Formula

$\mathrm{velocity}\left(\right(\mathrm{v})=\frac{\mathrm{distance}\left(\mathrm{d}\right)}{\mathrm{time}\left(\mathrm{t}\right)}$

$\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{speed}\mathrm{of}\mathrm{light}}$

(a) Calculate the time taken by radio signal to travel  150 km

Since the distance traveled by radio signal is $150\mathrm{km}$,

Thus, the distance will be,

localid="1663063329319" $\begin{array}{c}d=150\mathrm{km}\\ =150\times {10}^3\mathrm{m}\end{array}$

The time taken is given by

$\mathrm{t}=\frac{\mathrm{d}}{\mathrm{c}}........\left(\mathrm{i}\right)$

Here,

The speed of light, $3\times {10}^8\mathrm{m}/\mathrm{s}$

Substitute known values in the above equation.

$\mathrm{t}=\frac{15\times {10}^4}{3\times {10}^8}\mathrm{s}=5.0\times {10}^{-4}\mathrm{s}$

Hence, time taken by radio signal to travel $150\mathrm{km}$is $5.0\times {10}^{-4}\mathrm{s}$.

(b) Calculate the time taken by the sunlight to reflect from moon and enter our eye:

At full moon, the Sun and the Moon are on opposite sides of the Earth. So, the distance traveled by the sunlight will be the sum of distance traveled from the Sun to Moon and then from Moon to Earth.

Therefore, the total distance is

$d=d_{SM}+d_{ME}$

Here, $d_{SM}$is the distance between sun and moon, localid="1664199963775" $d_{ME}$is the distance between moon and earth.

Substitute known values in the above equation, and you obtain

$\begin{array}{rcl}\mathrm{d}& =& 1.5\times {10}^8\mathrm{km}+3.8\times {10}^5\mathrm{km}\\ & =& 1500\times {10}^5\mathrm{km}+3.8\times {10}^5\mathrm{km}\\ & =& 1503.8\times {10}^5\mathrm{km}\\ & =& 1.5\times {10}^{11}\mathrm{km}\end{array}$

Now, the time taken by the light to travel this distance is given by

$\begin{array}{rcl}\mathrm{t}& =& \frac{\mathrm{d}}{\mathrm{c}}\\ & =& \frac{1.51\times {10}^{11}\mathrm{m}}{3\times {10}^8\mathrm{m}/\mathrm{s}}\\ & =& 503.\\ & =& 8.4\min \end{array}$

Hence, the time taken by the sunlight to reflect from moon and enter our eye is $8.4\min$.

(c) Calculate the round-trip travel time for light between Earth and a spaceship orbiting Saturn

Distance between Earth and the spaceship is given as$1.3x{10}^9\mathrm{km}$.

Thus, the total distance for the roundtrip will be

role="math" localid="1663064483417" $\begin{array}{c}d=2(1.3\times {10}^9)\mathrm{km}\\ =2.6\times {10}^9\mathrm{km}\\ =2.6\times {10}^{12}\mathrm{m}\end{array}$

So, the time taken by for the roundtripbetween Earth and the spaceship is

$\begin{array}{rcl}\mathrm{t}& =& \frac{\mathrm{d}}{\mathrm{c}}\\ & =& \frac{2.6\times {10}^{12}\mathrm{m}}{3\times {10}^8\mathrm{m}/\mathrm{s}}\\ & =& 8.7\times {10}^3\mathrm{s}\\ & =& 2.4\mathrm{hours}\end{array}$

Hence, round-trip travel time for light between Earth and a spaceship orbiting Saturn is$2.4\mathrm{hours}$.

(d) Calculate the year in which the supernova explosion which resulted in Crab nebula occur

As given, the distance of Crab nebula is,

$\mathrm{d}=6500\mathrm{light}\mathrm{years}$$\mathrm{d}=6500\mathrm{light}\mathrm{years}$

And the speed of light is,

$\mathrm{c}=1.0\frac{\mathrm{light}\mathrm{year}}{\mathrm{year}}$

Since, the equation for time is,

$\begin{array}{rcl}\mathrm{t}& =& \frac{\mathrm{d}}{\mathrm{c}}\\ & =& \frac{6500\mathrm{light}\mathrm{year}}{1.0\mathrm{light}\mathrm{year}/\mathrm{year}}\\ & =& 6500\mathrm{year}\end{array}$

The explosion recorded by Chinese astronomers was in$1054\mathrm{AD}$.Thus, the explosion took place in the year is,

$1054-6550=-5466$

Hence, you can say that the explosion took place in year $5466\mathrm{BC}$.