Question

The mass of the Earth is $6 \times {10^{24}}$kg. The distance between the Earth and the Sun is $1.5 \times {10^{11}}$m. If the gravitational force between the two is $3.5 \times {10^2}N$, what is the mass of the sun?

Answer 1

We have the given values of,
Mass of the Earth is$6 \times {10^{24}}$kg,
The gravitational force between the two is$3.5 \times {10^2}N$,
The distance between the Earth and the Sun is $1.5 \times {10^{11}}$m
Here, we use the expression of Gravitational Force. The expression is
$F = \dfrac{{GMm}}{{{R^2}}}$
On putting all the given values in the expression we remain with a variable which is mass of sun so after simplifying the expression we get our answer.

Complete step by step solution:
Here, we use the expression of the Gravitational Force,
$F = \left( {\dfrac{{G{M_E}{M_S}}}{{{R^2}}}} \right)$
Here, F is the gravitational force
G is the gravitational constant
ME is the mass of the Earth
MS is the mass of the Sun
R is the distance between the Earth and the Sun
We know the value of the gravitational force, mass of the Earth, and the distance between the Earth and the Sun, and the value of the gravitational constant is $$6.67 \times {10^{ - 11}}$$$N{m^2}k{g^{ - 2}}$$
F = \left( {\dfrac{{G{M_E}{M_S}}}{{{R^2}}}} \right) \\
\Rightarrow {M_S} = \left( {\dfrac{{F \times {R^2}}}{{G \times {M_E}}}} \right) \\
$Now, put the value in above expression,$
{M_S} = \dfrac{{\left( {3.5 \times {{10}^2}} \right) \times {{\left( {1.5 \times {{10}^{11}}} \right)}^2}}}{{\left( {6.67 \times {{10}^{ - 11}}} \right) \times \left( {6 \times {{10}^{24}}} \right)}} \\
\Rightarrow {M_S} = \dfrac{{\left( {3.5 \times {{10}^2}} \right) \times \left( {2.25 \times {{10}^{22}}} \right)}}{{\left( {6.67 \times {{10}^{ - 11}}} \right) \times \left( {6 \times {{10}^{24}}} \right)}} \\
\Rightarrow {M_S} = \dfrac{{7.875 \times {{10}^{24}}}}{{40.02 \times {{10}^{13}}}} \\
\Rightarrow {M_S} = 0.19677 \times {10^{11}}kg \\
$**So, the mass of the sun is$0.19677 \times {10^{11}}$ kg**

Note: The term gravitational constant, it is the proportionality constant used in Newton’s Law of Universal Gravitation. It is represented by G. In Newton’s Law of the universal gravitation,
The F is the attractive force that is equal to the G (Gravitational Constant) times the product of the (M and m) masses and it’s divided by the square of the distance (r2) between them.
The g and G both are different, g represents the acceleration due to the gravity and the G is represented as the Gravitational Constant.
The value of g is $9.8m/{s^2}$ and the value of G is $6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$