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Gravitation — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsGravitation5 Marks
Question
Earth’s orbit is an ellipse with eccentricity $0.0167$. Thus, 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 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?
✓

Answer

According to the diagram,
$r_p$ = radius of perigee = $2R$
$r_a$ = radius of apogee = $6R$
a = semi - major axis of the ellipse

Hence, we can write
$r_a = a(1 + e) = 6R$
$P_p = a(1 - e) = 2R$
$\frac{\text{a(1+e)}}{\text{a(1}-\text{e})}=\frac{6\text{R}}{2\text{R}}=3$
By solving, we get eccentricity $\text{e}=\frac{1}{2}$
If $v_a$ and $v_p$ are the velocities of the satellite (of mass m) at aphelion and perihelion respectively, then by conservation of angular momentum
$\therefore\ \text{L}_\text{at perigee}=\text{L}_\text{at apogee}$
$\text{mv}_\text{p}\text{r}_\text{p}=\text{mv}_\text{a}\text{r}_\text{a}$
$\therefore\ \frac{\text{v}_\text{a}}{\text{v}_\text{p}}=\frac{\text{r}_\text{p}}{\text{r}_\text{a}}=\frac{1}{3}$
Applying conservation of energy,
Energy at perigee = Energy at apogee
where M is the mass of the earth
$\therefore\ \text{v}_\text{p}^2\Big(1-\frac{1}{9}\Big)=-2\text{GM}\Big(\frac{1}{\text{r}_\text{a}}-\frac{1}{\text{r}_\text{p}}\Big)$
$=2\text{GM}\Big(\frac{1}{\text{r}_\text{p}}-\frac{1}{\text{r}_\text{a}}\Big)$ $(\text{By putting v}_\text{a}=\frac{\text{v}_\text{p}}{3})$
$\text{v}_\text{p}=\frac{\Big[2\text{GM}\big(\frac{1}{\text{r}_\text{p}}-\frac{1}{\text{r}_\text{a}}\big)\Big]^{\frac{1}{2}}}{\Big[1-\big(\frac{\text{v}_\text{a}}{\text{v}_\text{p}}\big)^2\Big]^{\frac{1}{2}}}$
$=\Bigg[\frac{\frac{2\text{GM}}{\text{R}}\Big(\frac{1}{2}-\frac{1}{6}\Big)}{\Big(1-\frac{1}{9}\Big)}\Bigg]^{\frac{1}{2}}=\Big(\frac{\frac{2}{3}}{\frac{8}{9}}\frac{\text{GM}}{\text{R}}\Big)^{\frac{1}{2}}=\sqrt{\frac{3}{4}\frac{\text{GM}}{\text{R}}}=6.85\text{km/ s}$
$\text{v}_\text{p}=6.85\text{km/ s,}\text{ v}_\text{a}=2.28\text{km/ s,}$
For circular orbit of radius r,
$v_c$ = orbital velocity $=\sqrt{\frac{\text{GM}}{\text{r}}}$
For $r = 6R, v_c$ $=\sqrt{\frac{\text{GM}}{\text{6R}}}$ = 3.23km/ s

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