Question

If the sun radiates energy at the rate of 3.6 x 10 to power 33 ergs/sec the rate at which the sun is loosing mass is given by
(47 1 x 10 to power 12 gm/sec
(B) 8 x 10 to power 34 gm/sec
(C) 1 x 10 to pwer 23
(D) 7 x 10 to power 13 gm/sec​

Answer 1

Answer:

$\boxed{\mathfrak{Rate \ at \ which \ sun \ is \ loosing \ it's \ mass = 4 \times 10^{12} \ g/s}}$

Explanation:

Rate of energy radiated by sun ($\rm \dfrac{dE}{dt}$) = $\rm 3.6 \times 10^{33}$ ergs/sec

From mass-energy equivalence:

$\rm \implies E = mc^2 \\ \\ \sf Differentiating \ on \ both \ sides: \\ \rm \implies \dfrac{dE}{dt} = {c}^{2} \dfrac{dm}{dt} \\ \\ \rm \implies \dfrac{dm}{dt} = \dfrac{1}{ {c}^{2} } . \dfrac{dE}{dt}$

c → Speed of light ($\rm 3 \times 10^{10}$) cm/s

$\sf \dfrac{dm}{dt}$ → Rate at which sun is loosing it's mass

By substituting values in the equation we get:

$\rm \implies \dfrac{dm}{dt} = \dfrac{1}{ {(3 \times {10}^{10}) }^{2} } \times 3.6 \times {10}^{33} \\ \\ \rm \implies \dfrac{dm}{dt} = \dfrac{1}{ 9 \times {10}^{20} } \times 3.6 \times {10}^{33} \\ \\ \rm \implies \dfrac{dm}{dt} = 0.4 \times {10}^{33 - 20} \\ \\ \rm \implies \dfrac{dm}{dt} = 0.4 \times {10}^{13} \\ \\ \rm \implies \dfrac{dm}{dt} = 4 \times {10}^{12} \: g /s$

Answer 2

Answer:

Energy radiated per second =3.6×10

26

s

J

​

Mass changing into this energy per second =

c

2

E

​

=

(3×10

8

)

2

s

2

m

2

​

3.6×10

26

S

J

​

​

=4×10

9

s

kg

​