Question

Assume that the orbit of the earth is a circle of radius 150 \(\times\) 10\(^6\) km. The gravitational constant and the earth’s orbital velocity are given as 6.7 \(\times\) 10\(^{-11}\) N·m\(^2\)/kg\(^2\) and 30 \(\times\) 10\(^3\) m/s, respectively. The calculated mass of the sun is _________ \(\times\) 10\(^30\) kg (rounded off to two decimal places).

Hint:

The mass of the sun can be calculated using Newton's law of gravitation and centripetal force, considering the orbital velocity of the earth.

Answer 1

Correct Answer: 2

Step 1: Use Newton’s Law of Gravitation.
The gravitational force between two objects is given by: \[ F = \frac{G \cdot M_{\text{sun}} \cdot m_{\text{earth}}}{r^2}, \] where \(F\) is the force, \(G\) is the gravitational constant, \(M_{\text{sun}}\) is the mass of the sun, \(m_{\text{earth}}\) is the mass of the earth, and \(r\) is the radius of the earth's orbit. Using the centripetal force formula for an orbiting body: \[ F = m_{\text{earth}} \cdot v^2 / r, \] where \(v\) is the orbital velocity of the earth. Equating the two expressions for \(F\) and solving for \(M_{\text{sun}}\): \[ \frac{G \cdot M_{\text{sun}} \cdot m_{\text{earth}}}{r^2} = m_{\text{earth}} \cdot v^2 / r, \] \[ M_{\text{sun}} = \frac{v^2 \cdot r}{G}. \] Step 2: Calculate the mass of the sun.
Substituting the given values: \[ M_{\text{sun}} = \frac{(30 \times 10^3)^2 \cdot (150 \times 10^6 \times 10^3)}{6.7 \times 10^{-11}} = 1.99 \times 10^{30} \, \text{kg}. \] Step 3: Conclusion.
The mass of the sun is 1.99 \(\times\) 10\(^30\) kg.