91Ó°ÊÓ

First, let's find the gravitational force acting on 1 kilogram of water. We will use Newton's law of gravitation, which is given by: \( F = G\frac{m_1m_2}{r^2} \) Where: - \(F\) is the gravitational force (in Newtons) - \(G\) is the gravitational constant, approximately \(6.674 \times 10^{-11} \mathrm{Nm^2/kg^2}\) - \(m_1\) and \(m_2\) are the masses of the two objects (in kg) - \(r\) is the distance between the centers of the two objects (in m) In this case, \(m_1\) is the mass of the Sun (given as \(1.99 \times 10^{30} \mathrm{kg}\)) and \(m_2 = 1 \mathrm{kg}\) (the mass of water). The distance between the Earth and the Sun, \(r\), is given as \(1.50 \times 10^{11} \mathrm{m}\). Plug in the values and solve for the gravitational force: \( F = 6.674 \times 10^{-11} \frac{(1.99 \times 10^{30})(1)}{(1.50 \times 10^{11})^2} \)

The tidal force acting on a mass due to another mass is the difference in gravitational force acting on the closer side and the farther side. We can use the following formula to calculate the tidal force: \( T = 2 \times r \times \Delta F \) Where: - \(T\) is the tidal force (in Newtons) - \(r\) is the distance between the centers of the two objects (in m) - \(\Delta F\) is the difference in gravitational force on the closer side and farther side. Since we already calculated the gravitational force acting on the 1 kg of water (which we can assume is at the Earth's surface), we can use it to calculate the difference in gravitational force acting on the closer side and the farther side. Since the distance from the Sun to the Earth is much bigger than the Earth's radius, we can approximate \(\Delta F\) by the following: \(\Delta F = \frac{dF}{dr} \) Where \(\frac{dF}{dr}\) is the derivative of the gravitational force with respect to the distance \(r\). First, find the derivative of the gravitational force equation: \(\frac{dF}{dr} = G \frac{m_1 m_2}{r^3} \) Now, plug in the values of \(G\), \(m_1\), and \(r\) to find the value of \(\Delta F\): \(\Delta F = 6.674 \times 10^{-11} \frac{(1.99 \times 10^{30})}{(1.50 \times 10^{11})^3} \) And finally, calculate the tidal force using the formula for \(T\): \( T = 2 \times (1.50 \times 10^{11}) \times \Delta F \) When you perform these calculations, you will find the value of the Sun's tidal force on 1 kilogram of water.