Question

The energy from the sun reaches just outside the earth's atmosphere at a rate of 1400 W m−2. The distance between the sun and the earth is 1.5 × 1011 m.
(a) Calculate the rate which the sun is losing its mass.
(b) How long will the sun last assuming a constant decay at this rate? The present mass of the sun is 2 × 1030 kg.

Answer 1

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Given : r = 1.5\times 10^{11}r=1.5×10

11

m I = 1400I=1400 Wm^{-2}Wm

−2

m_o = 2\times 10^{30}m

o

=2×10

30

kg

Power radiated by sun P = I(4\pi r^2) = 1400\times (4\pi \times (1.5\times 10^{11})^2) = 3.96\times 10^{26}P=I(4πr

2

)=1400×(4π×(1.5×10

11

)

2

) =3.96×10

26

WW

Thus rate of losing mass \dfrac{dm}{dt} = \dfrac{P}{c^2} = \dfrac{3.96\times 10^{26}}{(3\times 10^8)^2} =4.4\times 10^9

dt

dm

=

c

2

P

=

(3×10

8

)

2

3.96×10

26

=4.4×10

9

kg/skg/s

Time for which the sun would last t = \dfrac{m_o}{\dfrac{dm}{dt}} =\dfrac{2\times 10^{30}}{4.4\times 10^9} = 4.55 \times 10^{20}t=

dt

dm

m

o

=

4.4×10

9

2×10

30

=4.55×10

20

s

\implies⟹ t = \dfrac{4.55\times 10^{20}}{365\times 86400} = 1.44\times 10^{13}t=

365×86400

4.55×10

20

=1.44×10

13

years

Answer 2

Explanation:

Given:

From the sun, the energy’s intensity, $I=1400 \mathrm{W} / \mathrm{m}^{2}$

Distance between Earth and the Sun, $R=1.5 \times 10^{11} \mathrm{m}$

$Power = Area x Intensity$

$P=A \times 1400$

$=1400 \times 4 \pi R^{2}$

$=1400 \times\left(1.5 \times 10^{11}\right)^{2} \times 4 \pi$

$=1400 \times 4 \pi \times 10^{22} \times(1.5)^{2}$

$Energy = Time \times Power$

In time t, the emitted energy, E = Pt

Sun has the mass that is used up to form this quantity of energy. So,

the loss in sun’s mass,

$\Delta m=E c 2$

$\Rightarrow \Delta m=p t c 2$ = $4.4 \times 109 \mathrm{kg} / \mathrm{s}$

So, at the rate of $4.4 \times 109 \mathrm{kg} / \mathrm{s}$, the sun is losing its weight.

(b) In 1 second, there is a $4.4 \times 10^{9} \mathrm{kg}$ loss.  

So, in $t^{\prime}=2 \times 10304.4 \times 109 \mathrm{s}, 2 \times 10^{30} \mathrm{kg}$ disintegrates  

$\Rightarrow \mathrm{t}^{\prime}=1.44 \times 1013 \mathrm{y}$