
How will you weight the sun, that is estimate its mass? The mean orbital of the earth around the sun is $1.5\times {{10}^{8}}\text{ km}\text{.}$
Answer
545.4k+ views
Hint: Estimate the period of one of its planets and the radius of planetary orbit.
Formula used:
\[G=\dfrac{4{{\pi }^{2}}{{r}^{3}}}{G{{T}^{2}}}\]
Where $M$ is mass, $r$ is radius of earth, $G$ is universal gravity constant, $T$ is time taken by earth to complete one revolution around the sun.
Complete step by step solution:
Given Data:
Orbital radius of earth $=1.5\times {{10}^{8}}\text{ km}$
$=1.5\times {{10}^{11}}\text{ m}$
Time taken by earth to complete one revolution around the Sun.
$T=365\text{ days}$
$=365\times 24\times 60\times 60s.$
$G=6.67\times {{10}^{11}}\text{ N}{{\text{m}}^{2}}\text{/k}{{\text{g}}^{2}}$
By formula, mass of Sun is
$$\[G=\dfrac{4{{\pi }^{2}}{{r}^{3}}}{G{{T}^{2}}}\]
$M=\dfrac{4\times {{3.14}^{2}}\times {{\left( 1.5\times {{10}^{11}} \right)}^{2}}}{6.67\times {{10}^{-11}}\times {{\left( 365\times 24\times 60\times 60 \right)}^{2}}}$
$M=2\times {{10}^{30}}\text{ kg}$
$\therefore $ mass of sun is $2\times {{10}^{30}}\text{ kg}$
Additional information:
Sun is the star at the centre of the solar system. It is a nearly perfect sphere of hot plasma heated to incandescence by nuclear fusion reaction in its core, radiating the energy mainly as visible light and infrared radiation. Different parts of the Sun rotate at different speeds as it is a great big sphere of hydrogen gas. The surface of the sun reaches a temperature of $6000$ kelvin. Above the surface of the sun is region of the atmosphere called the chromosphere where temperature can reach $100,000K$
Note:
Mean distance from Earth $1AU\approx 1.496\times {{10}^{4}}$
Volume $=1.41\times {{10}^{18}}\text{ k}{{\text{m}}^{3}}$
Mass $=1.9885\times {{10}^{30}}\text{ kg}$
Sun is the most important source of energy for life on earth. Its diameter $109$ times that of Earth.
The solar constant is the amount of power that the Sun deposits per unit area that is directly exposed to sunlight. The solar constant is equal to approximately $1368\text{ W/}{{\text{m}}^{2}}$ at a distance of one astronomical unit from the sun.
Formula used:
\[G=\dfrac{4{{\pi }^{2}}{{r}^{3}}}{G{{T}^{2}}}\]
Where $M$ is mass, $r$ is radius of earth, $G$ is universal gravity constant, $T$ is time taken by earth to complete one revolution around the sun.
Complete step by step solution:
Given Data:
Orbital radius of earth $=1.5\times {{10}^{8}}\text{ km}$
$=1.5\times {{10}^{11}}\text{ m}$
Time taken by earth to complete one revolution around the Sun.
$T=365\text{ days}$
$=365\times 24\times 60\times 60s.$
$G=6.67\times {{10}^{11}}\text{ N}{{\text{m}}^{2}}\text{/k}{{\text{g}}^{2}}$
By formula, mass of Sun is
$$\[G=\dfrac{4{{\pi }^{2}}{{r}^{3}}}{G{{T}^{2}}}\]
$M=\dfrac{4\times {{3.14}^{2}}\times {{\left( 1.5\times {{10}^{11}} \right)}^{2}}}{6.67\times {{10}^{-11}}\times {{\left( 365\times 24\times 60\times 60 \right)}^{2}}}$
$M=2\times {{10}^{30}}\text{ kg}$
$\therefore $ mass of sun is $2\times {{10}^{30}}\text{ kg}$
Additional information:
Sun is the star at the centre of the solar system. It is a nearly perfect sphere of hot plasma heated to incandescence by nuclear fusion reaction in its core, radiating the energy mainly as visible light and infrared radiation. Different parts of the Sun rotate at different speeds as it is a great big sphere of hydrogen gas. The surface of the sun reaches a temperature of $6000$ kelvin. Above the surface of the sun is region of the atmosphere called the chromosphere where temperature can reach $100,000K$
Note:
Mean distance from Earth $1AU\approx 1.496\times {{10}^{4}}$
Volume $=1.41\times {{10}^{18}}\text{ k}{{\text{m}}^{3}}$
Mass $=1.9885\times {{10}^{30}}\text{ kg}$
Sun is the most important source of energy for life on earth. Its diameter $109$ times that of Earth.
The solar constant is the amount of power that the Sun deposits per unit area that is directly exposed to sunlight. The solar constant is equal to approximately $1368\text{ W/}{{\text{m}}^{2}}$ at a distance of one astronomical unit from the sun.
Recently Updated Pages
Complete reduction of benzene diazonium chloride with class 12 chemistry CBSE

How can you identify optical isomers class 12 chemistry CBSE

The coating formed on the metals such as iron silver class 12 chemistry CBSE

Metals are refined by using different methods Which class 12 chemistry CBSE

What do you understand by denaturation of proteins class 12 chemistry CBSE

Assertion Nitrobenzene is used as a solvent in FriedelCrafts class 12 chemistry CBSE

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

RNA and DNA are chiral molecules their chirality is class 12 chemistry CBSE

