Question

Calculate the mass of the sun, assuming the Earth's orbit around the sun is circular, with radius r = 1.5x10 raise to power 8 km.

Answer 1

Answer on Question 55023, Physics, Mechanics | Kinematics | Dynamics

Question:

Calculate the mass of the Sun, assuming the Earth's orbit around the Sun is circular, with the radius $r = 1.5 \cdot 10^{8} \, \text{km}$.

Solution:

By the universal gravitation equation, the gravitational force acting from the Sun on the Earth looks like:

$$F_{SE} = G \frac{M_{Sun} m_{Earth}}{r^{2}},$$

where, $G$ is the gravitational constant, $M_{Sun}$ is the mass of the Sun, $m_{Earth}$ is the mass of the Earth and $r$ is the distance between the Sun and the Earth.

Since, we assuming the Earth's orbit around the Sun is circular, the gravitational force $F_{SE}$ must balance the centripetal force on the Earth:

$$F_{c} = \frac{m_{Earth} v^{2}}{r},$$

where, $m_{Earth}$ is the mass of the Earth, $v$ is the orbital speed of the Earth and $r$ is the radius of the Earth's orbit.

So, we can write:

$$F_{SE} = F_{c},$$ $$G \frac{M_{Sun} m_{Earth}}{r^{2}} = \frac{m_{Earth} v^{2}}{r},$$ $$M_{Sun} = v^{2} \frac{r}{G}.$$

Because the Earth travels around the entire circumference of the circle which is $2\pi r$ in the period $T$, this means that the orbital speed of the Earth must be $v = \frac{2\pi r}{T}$.

Substituting the expression for the orbital speed into the last equation we can calculate the mass of the Sun:

$$M_{Sun} = \frac{4\pi^2 r^3}{T^2 G} = \frac{4 \cdot 3.14^2 \cdot (1.5 \cdot 10^{11} m)^3}{(3.15 \cdot 10^7 s)^2 \cdot 6.67 \cdot 10^{-11} \frac{N m^2}{k g^2}} = 2.0 \cdot 10^{30} k g.$$

Answer:

$$M_{Sun} = 2.0 \cdot 10^{30} k g.$$

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