Question

A comet is in highly elliptical orbit around the sun. The period of the comet’s orbit is 90 days. Some statements are given regarding collision between the comet and the Earth. Mark the correct statement. [Mass of sun $= 2 \times {10^{30}}kg$, mean distance between the earth and the Sun $= 1.5 \times {10^{11}}m$]
(a) Collision is there
(b) Collision is not possible
(c) Collision may or may not be there
(d) Enough information is not given

Answer 1

Hint: It is given that the orbit of comet is highly elliptic, in order to get the actual path, we need the eccentricity of ellipse and major and minor axis calculations. But if we can’t calculate just with Time period data. Hence, we can try thinking by approximating the ellipse to be a circle with the same center and use Kepler law.
We can use Kepler’s third law which relates the time period of an axis to its radius (if orbit is circular) or to the semi major axis (if the orbit is elliptical). This will help us to check whether the orbit of the earth and the comet coincide with each other.

Formula used:
Kepler’s third law:
$T = 2\pi \sqrt {\dfrac{{{r^3}}}{{G{M_S}}}}$ …… (1)
where,
T is the time period of the motion of the body around the sun
r is the length of the semi-major axis of the ellipse (or radius in case of circular orbit)
G is the gravitational constant
${M_S}$is the mass of the Sun

Complete step-by-step answer:
Given:
1. Time period of comet’s orbit around the Sun $({T_C}) = 90days$
2. Mass of sun ${M_S} = 2 \times {10^{30}}kg$
3. Mean distance between the earth and the Sun ${r_{ES}} = 1.5 \times {10^{11}}m$

To find: Whether the length of the major axis of the comet’s orbit coincides with the mean distance between the earth and the Sun.

Step 1 of 3:
Square both sides of equation (1). This gives:
${T^2} = 4{\pi ^2}\dfrac{{{r^3}}}{{G{M_S}}}$ …… (2)
Rearrange equation (2) to give r:
$r = \sqrt[3]{{\dfrac{{{T^2}G{M_S}}}{{4{\pi ^2}}}}}$ …… (3)

Step 2 of 3:
Convert ${T_C}$ from days to seconds:
${T_C} = days \times 24hr \times 3600s \\\ {T_C} = 90 \times 24 \times 3600 \\\ {T_C} = 7776000s \\\$
Substitute the values of G, T and ${M_S}$ in eq 3:

r = \sqrt[3]{{\dfrac{{{{7776000}^2} \times 6.67 \times {{10}^{ - 11}} \times 2 \times {{10}^{30}}}}{{4{\pi ^2}}}}} \\\ r = 2.73 \times {10^7}m \\\ $$ …… (4) To know about r calculated above see additional information. Step 3 of 3: Let the length of the major axis of elliptical orbit be a. Find the value of a using the value of r obtained: $ a = 2r \\\ a = 2 \times 2.73 \times {10^7} \\\ a = 5.46 \times {10^7}m \\\ $ Compare with the mean distance between the earth and the Sun: $a < {r_{ES}}$ This implies that the orbit of the comet lies inside the orbit of the earth around the Sun. Hence the comet will never collide with earth. _Therefore option B is the correct answer._ Additional Information: This r calculated for the comet is the equivalent radius of the circle which has the same time period as ellipse hence length of major axis can’t be greater than 2r as max eccentricity could be 1. Note: In questions like these, use Kepler’s third law to find the value of radius (for circular orbit) or major axis (for elliptical orbit) of the orbit. Comparing the r values will determine whether the orbits lie in each other’s path or not. If the two Values of r are different, the 2 bodies will never collide.