91Ó°ÊÓ

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.

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

Formula

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

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

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

Thus, the distance will be,

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

The time taken is given by

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

Here,

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

Substitute known values in the above equation.

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

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

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

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

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

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

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

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" $\begin{array}{c}d=2(1.3×{10}^9)\mathrm{km}\\ =2.6×{10}^9\mathrm{km}\\ =2.6×{10}^{12}\mathrm{m}\end{array}$

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

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

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

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 $5466\mathrm{BC}$.