Question

At Earth's location, the intensity of sunlight is 1.5 kW / ${\mathbf{m}}^{\mathbf{2}}$. If no energy escaped Earth, by how much would Earth's mass increase in 1 day?

Short answer

The increase in Earth’s mass for no escape of Sun’s energy from Earth is $7.34\times {10}^5\mathrm{kg}$.

Step-by-step solution

Identification of given data

The intensity of sunlight on Earth is $\mathrm{I}=1.5\mathrm{kW}/{\mathrm{m}}^2$.

The number of seconds in a day on Earth is $\mathrm{t}=1\mathrm{day}=86400\mathrm{s}$.

The energy received by Earth per unit surface area from the sun is called the solar constant.

Determination of energy received by Earth in one day

The energy received by Earth in time t is given as:

$E=I.A.t$

Here, A is the surface area of Earth and its value is $5.10\times {10}^{14}{\mathrm{m}}^2$.

For the given values, equation becomes-

$$\begin{gathered} E=\left(1.5\mathrm{kW}/{\mathrm{m}}^2\right)\left(\frac{{10}^3\mathrm{J}/\mathrm{s}}{1\mathrm{kW}}\right)\left(5.10\times {10}^{14}\right)\left(86400\mathrm{s}\right) \\ E=6.61\times {10}^{22}\mathrm{J} \end{gathered}$$

Determination of increase in Earth’s mass

The increase in Earth’s mass is given as:

$E=m{c}^2$

Here, c is the speed of light in vacuum and its value is $3\times {10}^8\mathrm{m}/\mathrm{s}$

$$\begin{gathered} 6.61\times {10}^{22}\mathrm{J}=\mathrm{m}{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2 \\ m=7.34\times {10}^5\mathrm{kg} \end{gathered}$$

Therefore, the increase in Earth’s mass for no escape of energy from Earth is $7.34\times {10}^5\mathrm{kg}$.