Question

The mass of the Sun is $\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{30}}\mathbf{}\mathit{k}\mathit{g}$. The mass of the Earth is $\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}\mathbf{}\mathit{k}\mathit{g}\mathbf{.}$and their center-to-center distance is $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathit{m}$. Suppose that at some instant the Sun's momentum is zero (it's at rest). Ignoring all effects but that of the Earth, what will the Sun's speed be after one day? (Very small changes in the velocity of a star can be detected using the "Doppler" effect, a change in the frequency of the starlight, which has made it possible to identify the presence of planets in orbit around a star.)

Short answer

The value of force$F=35.573\times {10}^{21}N$.

Step-by-step solution

Identification of given data

Mass of sun$m_1=2\times {10}^{30}kg$

Mass of earth$m_2=6\times {10}^{24}kg$

Distance$1.5\times {10}^{11}m$

Calculation of the gravitational force

According to the Newton’s low of gravitation

According to the Newton’s low of gravitation

$F=\frac{G{m}_1{m}_2}{r^2}$

Where

G – Gravitational constant

$m_1$- Mass of 1 object

$m_2-$Mass of 2 objects

$r-$Distance between two object

So after putting the values we get

$$\begin{gathered} F=\frac{G{m}_1{m}_2}{r^2} \\ F=\frac{6.67\times {10}^{-11}{\mathrm{m}}^3/{\mathrm{kgs}}^2\times 2\times {10}^{30}\times 6\times {10}^{24}}{{\left(1.5\times {10}^{11}\right)}^2} \\ F=35.573\times {10}^{21}\mathrm{N} \end{gathered}$$

Thus, the value of force is $F=35.573\times {10}^{21}\mathrm{N}$