Earth's orbit is an ellipse with eccentricity 0.0167 . Thus, the earth's distance from the sun and speed as it moves around the sun varies from day-to-day. This means that the length of the solar day is not constant through the year. Assume that the earth's spin axis is normal to its orbital plane and find out the length of the shortest and the longest day. A day should be taken from noon to noon. Does this explain variation of length of the day during the year?
Consider the diagram. Let $m$ be the mass of the earth, $v_p, v_a$ be the velocity of the earth at perigee and apogee respectively. Similarly, $\omega_p$ and $\omega_a$ are corresponding angular velocities.
Angular momentum and areal velocity are constant as the earth orbits the sun.
At perigee, $r_p^2 \omega_p=r_a^2 \omega_a$ at apogee $$\quad$$ .... (i)
If $a$ is the semi-major axis of the earth's orbit, then $r_p=a(1-e)$ and $r_a=a(1+e)\quad \text{... (ii)}$
$$ \begin{array}{ll} \therefore \frac{\omega_p}{\omega_a}=\left(\frac{1+e}{1-e}\right)^2, e=0.0167 & \text { [from Eqs. (i) and (ii)] } \\ \therefore\frac{\omega_p}{\omega_a}=1.0691 & \end{array}$$
Let $\omega$ be angular speed which is geometric mean of $\omega_p$ and $\omega_a$ and corresponds to mean solar day,
$$\begin{array}{ll} \therefore & \left(\frac{\omega_p}{\omega}\right)\left(\frac{\omega}{\omega_a}\right)=1.0691 \\ \therefore & \frac{\omega_p}{\omega}=\frac{\omega}{\omega_a}=1.034 \end{array}$$
If $\omega$ corresponds to $1^{\circ}$ per day (mean angular speed), then $\omega_p=1.034^{\circ}$ per day and $\omega_{\mathrm{a}}=0.967^{\circ}$ per day. Since, $361^{\circ}=24$, mean solar day, we get $361.034^{\circ}$ which corresponds to $24 \mathrm{~h}, 8.14^{\prime \prime}$ ( $8.1^{\prime \prime}$ longer) and $360.967^{\circ}$ corresponds to 23 h 59 min $52^{\prime \prime}$ ( $7.9^{\prime \prime}$ smaller).
This does not explain the actual variation of the length of the day during the year.
A satellite is in an elliptic orbit around the earth with aphelion of $6 R$ and perihelion of $2 R$ where $R=6400 \mathrm{~km}$ is the radius of the earth. Find eccentricity of the orbit. Find the velocity of the satellite at apogee and perigee. What should be done if this satellite has to be transferred to a circular orbit of radius $6 R$ ? $$ \left[G=6.67 \times 10^{-11} \text { SI unit and } M=6 \times 10^{24} \mathrm{~kg}\right] $$
Given,
$r_p=$ radius of perihelion $=2 R$
$$r_a=\text { radius of apnelion }=6 R$$
Hence, we can write
$$ \begin{aligned} & r_a=a(1+e)=6 R \quad \text{... (i)}\\ & r_p=a(1-e)=2 R \quad \text{... (ii)} \end{aligned}$$
Solving Eqs. (i) and (ii), we get eccentricity, $e=\frac{1}{2}$
By conservation of angular momentum, angular momentum at perigee $=$ angular momentum at apogee
$$\begin{array}{ll} \therefore & m v_p r_p=m v_a r_a \\ \therefore & \frac{v_a}{v_p}=\frac{1}{3} \end{array}$$
where $m$ is mass of the satellite. Applying conservation of energy, energy at perigee $=$ energy at apogee
$$\frac{1}{2} m v_p^2-\frac{G M m}{r_p}=\frac{1}{2} m v_a^2-\frac{G M m}{r_a}$$
where $M$ is the mass of the earth.
$$\begin{aligned} &\begin{aligned} \therefore\quad v_p^2\left(1-\frac{1}{9}\right) & =-2 G M\left(\frac{1}{r_a}-\frac{1}{r_p}\right)=2 G M\left(\frac{1}{r_p}-\frac{1}{r_a}\right) \quad\left(\text { By putting } v_a=\frac{v_p}{3}\right) \\ v_p & =\frac{\left[2 G M\left(\frac{1}{r_p}-\frac{1}{r_a}\right)\right]^{1 / 2}}{\left[1-\left(V_a / V_p\right)^2\right]^{1 / 2}}=\left[\frac{\frac{2 G M}{R}\left(\frac{1}{2}-\frac{1}{6}\right)}{\left(1-\frac{1}{9}\right)}\right]^{1 / 2} \\ & =\left(\frac{2 / 3}{8 / 9} \frac{G M}{R}\right)^{1 / 2}=\sqrt{\frac{3}{4}} \frac{G M}{R}=6.85 \mathrm{~km} / \mathrm{s} \\ v_p & =6.85 \mathrm{~km} / \mathrm{s}, v_a=2.28 \mathrm{~km} / \mathrm{s} \end{aligned} \end{aligned}$$
For circular orbit of radius $r$,
$$\begin{aligned} v_c & =\text { orbital velocity }=\sqrt{\frac{G M}{r}} \\ \text{For}\quad r & =6 R, v_c=\sqrt{\frac{G M}{6 R}}=3.23 \mathrm{~km} / \mathrm{s} \end{aligned}$$
Hence, to transfer to a circular orbit at apogee, we have to boost the velocity by $\Delta=(3.23-2.28)=0.95 \mathrm{~km} / \mathrm{s}$. This can be done by suitably firing rockets from the satellite.