Question

How will you weight the sun that estimates its mass? The mean orbital radius of the earth around the sun is $1.5 \times {10^{11}}m$.

Answer 1

Hint: The earth revolves around the sun along an orbital path. The earth to keep its circular motion, it needs a centripetal force that will keep it in orbit. Here, the gravitational force(between sun and earth) supplies the required centripetal force.

Formula Used:
The object of mass m at a distance $R$ from a body of mass $M$, feels a gravitational force
${F_G} = \dfrac{{GMm}}{{{R^2}}}$
where, $G$ is the universal gravitational constant.
For an object of mass $m$ rotating with velocity $v$ along a circular path of radius $R$ obtains a centripetal force
${F_c} = \dfrac{{m{v^2}}}{R}$

Complete step by step answer:
Given:
The mean orbital radius of the earth around the sun is $1.5 \times {10^{11}}m$.
To get: The mass of the sun.
Step 1:
Let the mass of the sun is ${M_s}$, the mass of the earth is ${M_e}$, the distance between them that is the orbital radius is $R = 1.5 \times {10^{11}}m$.
Hence, represent the gravitational force acting on the earth ${F_G}$ from eq (1) with current variables.
${F_G} = \dfrac{{G{M_s}{M_e}}}{{{R^2}}}$
Now calculate the centripetal force on earth ${F_c}$ from eq (2)
${F_c} = \dfrac{{{M_e}{v_e}^2}}{R}$
where, ${v_e}$ is the velocity of earth.
Step 2:
Now as the gravitational force ${F_G}$ supplies the required centripetal force ${F_c}$, so they are equal.
Equate the eq (3) and eq (4)
$
  {F_G} = {F_c} \\
   \Rightarrow \dfrac{{G{M_s}{M_e}}}{{{R^2}}} = \dfrac{{{M_e}{v_e}^2}}{R} \\
 $
Hence rewrite the relation to get an expression for the mass of the sun ${M_s}$.
${M_s} = \dfrac{{{v_e}^2R}}{G}$
Step 3:
The orbital radius of the earth around the sun is $R = 1.5 \times {10^{11}}m$.
Hence, for one complete orbit, the earth traverses a path of the circumference of the orbit, that is $2\pi R$.
The earth takes 1 year to complete an orbit around the sun.
So, the time period is
 $
  T = 365 \times 24 \times 60 \times 60s \\
   \Rightarrow T = 3.2 \times {10^7}s \\
 $
Hence, the velocity of the earth ${v_e}$ is found to be
$
  {v_e} = \dfrac{{2\pi R}}{T} \\
   \Rightarrow {v_e} = \dfrac{{2 \times 3.14 \times 1.5 \times {{10}^{11}}}}{{3.2 \times {{10}^7}}}m.{s^{ - 1}} \\
 $
Step 4:
The value of the universal gravitational constant is $G = 6.67 \times {10^{ - 11}}N.{m^2}.k{g^{ - 2}}$.
Calculate the mass of the sun ${M_s}$ from eq (5) by putting all the required values
$
  {M_s} = \dfrac{{{v_e}^2R}}{G} \\
   \Rightarrow {M_s} = \dfrac{{{{\left( {\dfrac{{2 \times 3.14 \times 1.5 \times {{10}^{11}}}}{{3.2 \times {{10}^7}}}} \right)}^2} \times 1.5 \times {{10}^{11}}}}{{6.67 \times {{10}^{ - 11}}}}kg \\
   \Rightarrow {M_s} = \dfrac{{{{\left( {2.94 \times {{10}^4}} \right)}^2} \times 1.5 \times {{10}^{22}}}}{{6.67}}kg \\
   \Rightarrow {M_s} = 8.64 \times {10^8} \times 0.22 \times {10^{22}}kg \\
   \Rightarrow {M_s} = 1.9 \times {10^{30}}kg \\
 $
$\therefore $ The mass of the sun ${M_s}$ is $1.9 \times {10^{30}}kg$.
Final Answer:
Thus by using the laws of gravitation and the simple newtonian mechanics you can estimate the weight of the sun rather the mass of the sun as ${M_s} = 1.9 \times {10^{30}}kg$.

Note: You can consider the orbit of the earth around the sun as a circular one. You should be careful with units while putting the values. Though $G$ is universal gravitational constant it is not just a number it has a prominent unit corresponding to its value. You should be careful while using it. Also note that all the units are in a particular system for example S.I units in this case. Here we are neglecting the interaction of earth with other planets(masses) in order to calculate the centripetal force on it.