91Ó°ÊÓ

First, let's find the total power (energy per unit time) emitted by the Sun, using the given intensity and the surface area of a sphere. The intensity (I) of the solar energy is given as \(1.4\, \mathrm{kW/m^2}\) and the distance (r) from the center of the Sun to the Earth is approximately \(1.5 \times 10^{11}\ \mathrm{m}\). We can use the surface area of a sphere (\(4\pi r^2\)) to calculate the total power emitted by the Sun: Power (P) = Intensity (I) × Surface Area (A) P = \(1.4\,\mathrm{kW/m^2} \times 4\pi(1.5 \times 10^{11}\,\mathrm{m})^2\)

Remember that 1 kW equals 1000 W. So, we need to convert the power from kilowatts to watts. P = \((1.4 \times 10^3)\,\mathrm{W/m^2} \times 4\pi(1.5 \times 10^{11}\,\mathrm{m})^2\)

Now, solve for the power: P = \(6.366 \times 10^{26}\,\mathrm{W}\)

To find the mass loss per day, we need to convert power to energy and use the energy-mass equivalence formula (E = mc²), where E is energy, m is mass, and c is the speed of light (\(3 \times 10^8\ \mathrm{m/s}\)). Energy per day (E) = Power (P) × Time (one day in seconds) E = \(6.366 \times 10^{26}\,\mathrm{W} \times 24\,\mathrm{h/day} \times 60\,\mathrm{min/h} \times 60\,\mathrm{s/min}\) E = \(5.496 \times 10^{31}\,\mathrm{J}\) (Joules) Now, we can use the energy-mass equivalence formula to find the mass loss (m): m = \(\frac{E}{c^2}\) m = \(\frac{5.496 \times 10^{31}\,\mathrm{J}}{(3 \times 10^8\,\mathrm{m/s})^2}\) m = \(6.104 \times 10^{11}\,\mathrm{kg}\) So, the Sun loses \(6.104 \times 10^{11}\,\mathrm{kg}\) of mass per day.

Now, let's find the percentage of the Sun's mass that is lost per day. The mass (M) of the Sun is approximately \(1.989 \times 10^{30}\,\mathrm{kg}\). Mass loss percentage = \(\frac{mass\, loss\, per\, day}{total\, mass\, of\, the\, Sun} \times 100\%\) Percentage = \(\frac{6.104 \times 10^{11}\,\mathrm{kg}}{1.989 \times 10^{30}\,\mathrm{kg}} \times 100\%\) Percentage ≈ \(3.067 \times 10^{-6}\%\) Thus, the Sun loses about \(3.067 \times 10^{-6}\%\) of its mass per day.