Chapter 26, Problem 57GP

(a)

To determine

The rate of mass loss.

Answer to Problem 57GP

  $\text{4}\text{.4}{\text{×10}}^{\text{9}}\text{kg/s}$ .

Explanation of Solution

Given:

The rate of energy radiated by the sun is, $\text{P}\text{=}{\text{4×10}}^{\text{26}}\text{W}$

Formula used:

The relation between the power and the energy is: $\text{P}\text{=}\frac{\text{E}}{\text{Δt}}$

  $\text{P}\text{=}\frac{{\text{m}}_{\text{0}}{\text{c}}^2}{\text{Δt}}$ or $\frac{{\text{m}}_{\text{0}}}{\text{Δt}}\text{=}\frac{\text{P}}{{\text{c}}^{\text{2}}}$

Where,

  $P$ is power (rate of energy)

  $E$ is energy

  $\Delta t$ is time dilation

  $m_0$ is rest mass

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

Calculation:

By substituting the values in the formula: $\frac{{\text{m}}_{\text{0}}}{\text{Δt}}\text{=}\frac{\text{P}}{{\text{c}}^{\text{2}}}$

  $\frac{{\text{m}}_{\text{0}}}{\text{Δt}}\text{=}\frac{\left({\text{4×10}}^{\text{26}}\text{W}\right)}{{\left(\text{3}\text{.00}\text{×}{\text{10}}^{\text{8}}\text{m/s}\right)}^{\text{2}}}\text{=4}{\text{.4×10}}^{\text{9}}\text{kg/s}$

(b)

To determine

The time taken to loss the mass.

Answer to Problem 57GP

  $\text{4}\text{.3}{\text{×10}}^{\text{7}}\text{yr}$ .

Explanation of Solution

Given:

The mass of earth is $\text{5}\text{.98}{\text{×10}}^{\text{24}}\text{kg}$

The rate of mass loss is $\text{4}\text{.4}{\text{×10}}^{\text{9}}\text{kg/s}$ .

Formula used:

It is known that, $\text{Δt=}\frac{\Delta \text{m}}{\text{rate}}$

Where,

  $\Delta t$ is time dilation

  $m$ is mass

Calculation:

By substituting the values in the formula: $\text{Δt=}\frac{\Delta \text{m}}{\text{rate}}$

  $\text{Δt =}\frac{\left(\text{5}{\text{.98×10}}^{\text{24}}\text{kg}\right)}{\left(\text{4}{\text{.4×10}}^{\text{9}}\text{kg/s}\right)\left(\text{3}{\text{.16×10}}^{\text{7}}\text{s/yr}\right)}\text{= 4}{\text{.3×10}}^{\text{7}}\text{yr}$

(c)

To determine

The time for which the Sun could last.

Answer to Problem 57GP

  $\text{1}\text{.4}{\text{×10}}^{\text{13}}\text{yr}$ .

Explanation of Solution

Given:

The mass of sun is $\text{2}\text{.0}{\text{×10}}^{\text{30}}\text{kg}$

The rate of mass loss is $\text{4}\text{.4}{\text{×10}}^{\text{9}}\text{kg/s}$ .

Formula used:

It is known that, $\text{Δt=}\frac{\Delta \text{m}}{\text{rate}}$

Where,

  $\Delta t$ is time dilation

  $m$ is mass

Calculation:

By substituting the values in the formula: $\text{Δt=}\frac{\Delta \text{m}}{\text{rate}}$

  $\text{Δt =}\frac{\left(\text{2}{\text{.0×10}}^{\text{30}}\text{kg}\right)}{\left(\text{4}{\text{.4×10}}^{\text{9}}\text{kg/s}\right)\left(\text{3}{\text{.16×10}}^{\text{7}}\text{s/yr}\right)}\text{= 1}{\text{.4×10}}^{\text{13}}\text{yr}$