Motion Of Planets And Satellites
WBBSE Class 11 Motion of Planets and Satellites Notes
Newton’s law of gravitation can be applied to arrive at Kepler’s laws theoretically. Using Newton’s law of gravitation, it can also be shown that the revolution of an object, natural or artificial (like a satellite), around another object (say, a planet) is similar to the motion of planets revolving around the sun.
- This means that satellites also follow Kepler’s laws within the parameter that the mass of the revolving object should be negligible compared to the mass of the object around which it revolves as in the case of the solar system.
- It should be noted that the eccentricities of the orbits of the planets in the solar system are very law, i.e., the orbits are almost circular. As a circle is a special case of ellipse (for circles, e = 0 ), Kepler’s laws are equally applicable for circular orbits. For the case of calculation, in the following cases, the orbits are taken as circular.
- All of Kepler’s laws are valid for circular and elliptical orbits, i.e., for orbits with 0 ≤ e < 1. As parabolas and hyperbolas are open conic sections, neither Kepler’s first law nor his third law applies to them. For example, a comet that makes a single pass by the sun follows a parabolic or hyperbolic path.
Orbital Speed And Period Of Revolution Of Planets: Let a planet of mass m revolve in a circular orbit of radius r, with the sun of mass M0 at the centre of the orbit.
Orbital Speed: The speed With which a planet moves in its orbit is called its orbital speed(v). In the case of a circular orbit, the value of this orbital speed is a constant. The centripetal force needed for this circular motion = \(\frac{m v^2}{r}\).
Read and Learn More: Class 11 Physics Notes
The mutual gravitational force of attraction between a planet and the sun provides this centripetal force. GM0m The force of gravitation, in this case, is \(\frac{G M m}{r^2}\).
Hence, \(\frac{m v^2}{r}=\frac{G M_0 m}{r^2}\)
or, \(v^2=\frac{G M_0}{r}\)
or, \(v=\sqrt{\frac{G M_0}{r}}\)
It should be noted that in the case of an elliptical orbit orbital speed (v) is not constant. If the distance of a planet from the sun is r, at an instance v ∝ \(\frac{1}{r}\).
Understanding Kepler’s Laws of Planetary Motion
Orbital Angular Velocity: The angular velocity of the planet revolving in its orbit is called its orbital angular velocity (ω). This is related to orbital speed (v) as ω = \(\frac{v}{r}\). This ω varies in an elliptical orbit as r and v change, but in a circular orbit, the value of ω is a constant.
From equation (1) ω = \(\frac{\nu}{r}=\sqrt{\frac{G M_0}{r^3}}\)….(2)
Time Period Of Revolution: The time taken by a planet to revolve once around the sun is called its time period of revolution (T).
In time T, the planet performs one complete revolution around the sun and thus moves through a distance 2πr.
Hence, orbital speed \(v=\frac{2 \pi r}{T}\)
or, \(T=\frac{2 \pi r}{v}=2 \pi r \sqrt{\frac{r}{G M_0}}\) [from equation (1)]
∴ T = \(2 \pi \sqrt{\frac{r^3}{G M_0}}\)….(3)
Hence, \(T^2=4 \pi^2 \frac{r^3}{G M_0}\)
or, \(T^2= constant \times r^3\)….(4)
Orbital Motion of Satellites Explained
This is Kepler’s third law. From this law, it can be stated that the orbital time period is longer for the planets which are farther from the sun.
Mercury, the planet nearest to the sun, revolves around the sun in 88 days, whereas Neptune, the farthest planet in the solar family, revolves in 165 years. The International Astronomical Union has classified Pluto as a dwarf planet and excluded it from the solar planet family.
It can be noted from equations (1), (2) and (3) that the orbital speed, the orbital angular velocity and the time period of revolution of a planet do not depend on the mass of the planet m, though each quantity depends on the mass of the sun M0.
Hence, from the knowledge of distance r of a planet from the sun and its time period T, the mass of the sun M0 can be calculated using equation (3), but the mass of the planet m cannot be determined.
| Class 11 Physics | Class 12 Maths | Class 11 Chemistry |
| NEET Foundation | Class 12 Physics | NEET Physics |
Clearly, the above discussion is not only applicable to planets moving around the sun, but also applicable to any object orbiting the sun, or even in the case of satellites moving around their respective planets.
Motion Of Planets And Satellites Numerical Examples
Short Answer Questions on Planetary Motion
Example 1. The distance of the dwarf planet Pluto from the sun I is 40 times the average distance of the Earth from the sun. How much time does Pluto take to revolve around the sun?
Solution:
Given
The distance of the dwarf planet Pluto from the sun I is 40 times the average distance of the Earth from the sun.
Let T1 = time period of Pluto,
T2 = time period of the earth,
r1 = distance of Pluto from the sun and
r2 = distance of the earth from the sun
From Kepler’s law, \(\frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3} \text { or, } \frac{T_1^2}{(1)^2}=\left(\frac{40 r_2}{r_2}\right)^3=40^3\)
∴ \(T_1^2=40^3 \text { or, } T_1=253 \mathrm{y} \text { (approx.). }\)
Example 2. Assuming that the earth moves around the sun in a circular orbit, find the orbital speed (in km · h-1) and the orbital angular velocity of the earth. The radius of the earth’s orbit = 1.5 x 108 km.
Solution:
Given
Assuming that the earth moves around the sun in a circular orbit,
The radius of the earth’s orbit = 1.5 x 108 km.
Time period of the earth ( T) = 365 d = 365 x 24 h
The radius of the orbit (r) = 1.5 x 108 km
∴ Orbital speed \((\nu)=\frac{2 \pi r}{T}=\frac{2 \times \pi \times 1.5 \times 10^8}{365 \times 24}\)
= \(1.076 \times 10^5 \mathrm{~km} \cdot \mathrm{h}^{-1} \)
Orbital angular velocity \((\omega)=\frac{\nu}{r}=\frac{2 \pi}{T}=\frac{2 \times \pi}{365 \times 24}\)
= \(7.17 \times 10^{-4} \mathrm{rad} \cdot \mathrm{h}^{-1} \text {. }\)
Example 3. If the distance of the earth from the sun suddenly decreases to half its present value, then how many days will make a year?
Solution:
Let the present distance of the earth from the sun = r, time period (T) = 365 d.
If distance decreases to \(\frac{r}{2}\), the corresponding time period is T1(say).
Following Kepler’s law, \(\frac{T^2}{r^3}=\frac{T_1^2}{r_1^3} \text { or, } T_1^2=T^2 \cdot \frac{r_1^3}{r^3}\)
or, \(T_1=T \sqrt{\frac{r_1^3}{r^3}}\) = \(365 \sqrt{\frac{\left(\frac{r}{2}\right)^3}{r^3}}\) = \(365 \sqrt{\frac{1}{8}}=\frac{365}{2 \sqrt{2}}=129 \mathrm{~d}\)
Example 4. The mean orbital radius of the Earth around the sun is 1.5 x 108 km. Calculate the mass of the sun if G = 6.67 x 10-11 N · m² · kg-2.
Solution:
Given
The mean orbital radius of the Earth around the sun is 1.5 x 108 km.
Here r = 1.5 x 108 km = 1.5 x 1011 m
T = 365 days = 365 x 24 x 3600 s
Since, centripetal force required = gravitational force between the earth and the sun.
∴ \(\frac{m v^2}{r}=\frac{G M m}{r^2} \text { or, } \frac{m}{r}\left(\frac{2 \pi r}{T}\right)^2=\frac{G M m}{r^2}\)
or, \(M=\frac{4 \pi^2 r^3}{G T^2}=\frac{4 \times 9.87 \times\left(1.5 \times 10^{11}\right)^3}{6.67 \times 10^{-11} \times(365 \times 24 \times 3600)^2}\)
∴ M = \(2.01 \times 10^{30} \mathrm{~kg}\)
Some Useful Data On The Sun, The Earth And The Moon:
Mass Of The Sun: Let the mass of the sun = M1, the distance of a planet from the sun = r and the time period of the planet around the sun = T.
∴ \(M_0=\frac{4 \pi^2 r^3}{G T^2}\)….(1)
In the case of the earth, the average distance between the sun and the earth r = 1.5 x 108 km (approx.) = 15 x 1010 m. The time period of revolution of the earth = 365 d (approx.) = 365 x 24 x 3600 s
Substituting the values in equation (1), \(M_0=\frac{4 \times \pi^2 \times\left(15 \times 10^{10}\right)^3}{6.7 \times 10^{-11} \times(365 \times 24 \times 3600)^2}\)
= \(2 \times 10^{30} \mathrm{~kg} \text { (approx.) }\)
Mass Of The Earth: Mass of the earth can be obtained when the motion of the moon around the earth is considered. The average distance between the moon and the earth, r = 3.84 x 108 m
The time period of revolution of the moon, T = 27.3 d = 27.3 x 24 x 60 x 60 s
Hence, from equation (1), the mass of the earth
M = \(\frac{4 \times \pi^2 \times\left(3.84 \times 10^8\right)^3}{6.7 \times 10^{-11} \times(27.3 \times 24 \times 3600)^2}\)
= \(6 \times 10^{24} \mathrm{~kg} \text { (approx.) }\)
The mass of the earth can be calculated following an easier method if the value of the acceleration is due to gravity.
Gravitational Force and Planetary Motion
Gravitational Attraction Between The Sun And The Earth: The mass of the sun, M0 = 2 x 1030 kg; the mass of the earth, M = 6 x 1024 kg; the average distance between the sun and the earth, r = 15 x 1010 m.
Hence, the mutual force of attraction between the sun and the earth,
F = \(\frac{G M_0 M}{r^2}\)
= \(\frac{6.67 \times 10^{-11} \times\left(2 \times 10^{30}\right) \times\left(6 \times 10^{24}\right)}{\left(15 \times 10^{10}\right)^2}\)
= \(3.56 \times 10^{22} \mathrm{~N} .\)
Gravitational Attraction Between The Earth And The Moon: The mass of the Earth, M = 6 x 1024 kg; the mass of the Moon, m = 7.33 x 1022 kg; the average distance between the Earth and the Moon, r = 3.84 x 108 m.
Hence, the gravitational force of attraction between the two,
F = \(\frac{G M m}{r^2}\)
= \(\frac{6.67 \times 10^{-11} \times\left(6 \times 10^{24}\right) \times\left(7.33 \times 10^{22}\right)}{\left(3.84 \times 10^8\right)^2}\)
= \(1.99 \times 10^{20} \mathrm{~N}\)
Hence, \(\frac{\text{gravitational force between the sun and the earth}}{\text{gravitational force between theearth and the moon}}\)
= \(\frac{3.56 \times 10^{22} \mathrm{~N}}{1.99 \times 10^{20} \mathrm{~N}}=179 .\)