Question

The maximum distance from the Earth to the Sun (at aphelion) is $\mathbf{1}\mathbf{.}\mathbf{521}\mathbf{\times }\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathit{m}$, and the distance of closest approach (at perihelion) is $\mathbf{1}\mathbf{.}\mathbf{471}\mathbf{\times }\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathit{m}$. The Earth’s orbital speed at perihelion is $\mathbf{3}\mathbf{.}\mathbf{027}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{m}\mathbf{/}\mathit{s}$. Determine (a) the Earth’s orbital speed at aphelion and the kinetic and potential energies of the Earth–Sun system (b) at perihelion and (c) at aphelion. (d) Is the total energy of the system constant? Explain. Ignore the effect of the Moon and other planets.

Short answer

(a) The aphelian velocity is$2.93\times {10}^{4}m/s$.

Step-by-step solution

Given information

$$\begin{gathered} perihelionvelocity{v}_p=3.027\times {10}^{4}m/s \\ perihelionradius{r}_p=1.471\times {10}^{11}m/s \\ aphelionradius{r}_a=1.521\times {10}^{11}m/s \end{gathered}$$

Momentum conservation

The momentum conservation law states that the total momentum acting on eachother of two or more moving object body that is product of mass radius and velocity will be constant until and unless any external force does not get applied on it.

(a) Earth’s orbital speed at aphelion 

The exerted on the earth is calculated as zero so the generated angular momentum of the earth is conserved. By the conserved angular momentum we will calculate the speed at aphelion:

$$\begin{gathered} m{r}_a{v}_a=m{r}_p{v}_p \\ {v}_a=v_p\left(\frac{r_p}{r_a}\right) \\ {v}_a=3.027\times {10}^{4}m/s\left(\frac{1.471}{1.521}\right) \\ {v}_a=2.93\times {10}^{4}m/s \end{gathered}$$

The aphelian velocity is $2.93\times {10}^{4}m/s$