Question: The orbit of Earth around the Sun is almost circular. The closest and farthest distances are $\mathbf{1}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{ }\mathit{k}\mathit{m}$and $\mathbf{1}\mathbf{.}\mathbf{52}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathit{k}\mathit{m}$respectively. Determine the corresponding variations in (a) total energy, (b) gravitational potential energy, (c) kinetic energy, and (d) orbital speed. (Hint:Use conservation of energy and conservation of angular momentum.)

The distance of aphelion or farthest distance is$R_a=1.52\times {10}^{11}m.$

The distance of perihelion or closest distance is$R_p=1.47\times {10}^{11}m.$

According to the rule of conservation of energy, energy can only be transformed from one form of energy to another and cannot be created or destroyed.

A spinning system's conservation of angular momentum ensures that its spin stays constant unless it is perturbed by an outside force.

We can use the law of conservation of energy and the law of conservation of angular momentum to find the variation in the total energy. Then finding the P.E, K.E, and orbital speed at the perihelion and aphelion, we can find the variation in respective quantities at the perihelion and aphelion.

Formula:

$$\begin{gathered} \frac{1}{2}m{v}_a^2-\frac{G{M}_s{M}_E}{R_a}=\frac{1}{2}m{v}_p^2-\frac{G{M}_s{M}_E}{R_p} \\ {v}_a{R}_a=v_p{R}_p \end{gathered}$$

Where, $M_s$is the mass of sun, $M_E$is the mass of earth, V_ais the velocity of object at aphelion and v_pis the velocity of object at perihelion

We have the equation of conservation of energy as

$$\begin{gathered} T.E_a=T.E_P \\ \frac{1}{2}m{v}_a^2-\frac{G{M}_s{M}_E}{R_a}=\frac{1}{2}m{V}_p^2-\frac{G{M}_s{M}_E}{R_p} \end{gathered}$$

And the equation for conservation of angular momentum implies

$v_a{R}_a=v_p{R}_p$

The total energy is conserved, so there is no variation_.

E = 0

Variation in total energy at the perihelion and aphelion,

The difference in potential energy can be expressed as

$$\begin{gathered} U=U_a-U_p \\ =-\frac{G{M}_s{M}_E}{R_a}-\left(-\frac{G{M}_s{M}_E}{R_p}\right). \end{gathered}$$

$$\begin{gathered} U=-\left(6.67\times {10}^{-11}\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\left(1.99\times {10}^{30} \text{kg}\right)\left(5.98\times {10}^{24} \text{kg}\right)\left(\frac{1}{1.52\times {10}^{11} \text{m}}-\frac{1}{1.47\times {10}^{11} \text{m}}\right) \\ U\cong 1.8\times {10}^{32}\text{J}. \end{gathered}$$

Variation in potential energy at the perihelion and aphelion, $U=1.8\times {10}^{32}\text{J}.$

According to the principle of energy conservation,

$$\begin{gathered} K_a-K_p=U_a-U_p \\ K=1.8\times {10}^{32}J. \end{gathered}$$

The equation for change in kinetic energy is

$$\begin{gathered} K=K_a-K_p \\ =\frac{1}{2}{M}_E\left(v_a^2-v_p^2\right) \end{gathered}$$

Now, using the conservation of momentum equation,

$v_a{R}_a=v_p{R}_p$

We can write

$v_p=\frac{v_a{R}_a}{R_p}$

So, substituting this equation in the equation of change in kinetic energy, we get

$K=\frac{1}{2}{M}_E{v}_a^2\left(1-\frac{R_a^2}{R_p^2}\right)$

Using this, we can derive equation foras V_a,

$v_a=\sqrt{\frac{2K}{M_E\left(1-\frac{R_a^2}{R_p^2}\right)}}$

By solving this equation, we get

$v_a=2.95\times {10}^4\frac{\text{m}}{\text{s}}$

Now, the variation in speed is

_{$$\begin{gathered} v=v_p-v_a \\ =\left(1-\frac{R_p}{R_a}\right)v_a \end{gathered}$$}

$$\begin{gathered} v=\left(1-\frac{1.47\times {10}^{11} \text{m}}{1.52\times {10}^{11} \text{m}}\right)\left(2.95\times {10}^4 \frac{\text{m}}{\text{s}}\right) \\ v=-0.99\times {10}^3\frac{\text{m}}{\text{s}} \\ =-0.99\text{km/s} \end{gathered}$$

Variation in orbital speed at the perihelion and aphelion is$-0.99\text{km/s}$