Question

What is the magnitude of the gravitational force that acts between the Sun and the Earth? (Include a diagram)
How fast does the Earth need to be moving to stay in orbit around the sun?

Answer 1

1. According to the Newton's law of universal gravitation, the magnitude of the gravitational force that acts between the Sun and the Earth is given by the following expression:

$$F = G\cdot \dfrac{M_{Earth}M_{Sun}}{R^2}$$

where $M_{Earth} = 5.97\cdot 10^{24} kg$ and $M_{Sun} = 1.99 \cdot 10^{30} kg$ are masses of Earth and Sun respectively, $R = 150\cdot 10^9 m$ is the average distance between Earth and Sun and $G = 6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant.

Thus:

$$F = 6.67\cdot 10^{-11} \cdot \dfrac{5.97\cdot 10^{24}\cdot1.99 \cdot 10^{30}}{150^2\cdot 10^{18}} = 3.5\cdot 10^{22}N$$

2. According to the Newton's second law, the acceleration of Earth is:

$$a = F/M_{Earth} = \dfrac{ 3.5\cdot 10^{22}}{5.97\cdot 10^{24}} = 5.86\cdot 10^{-3} m/s^2$$

Let's assume, that Earth moves on a circular orbit with a constant speed. Then the speed will be:

$$v = \sqrt{aR} = \sqrt{5.86\cdot10^{-3}\cdot150\cdot10^{9}} = 2.96\cdot 10^4 m/s = 29.8 km/s$$

Answer. F = 3.5*10^22 N, v = 29.8 km/s