Physics |
Chemistry |
Biology |
Mathematics |

Science > Physics > Gravitation > You are Here |

- In the second century A.D., Ptolemy proposed Geocentric theory of the universe. According to this theory, the Earth is the centre of the universe and all other planets, sun and other stars revolve around the Earth.
- Indian astronomer, mathematician and philosopher Aryabhatta in 498 A.D. proposed that not only the Earth revolves around the sun but it rotates around its own axis.
- Polish astronomer Nicolaus Copernicus (1473 –1543) proposed the Heliocentric theory of the universe. According to this theory, the Sun is the centre of the galaxy and all other planets revolve around the Sun.
- The Danish astronomer Tyco Brahe made the most accurate observations possible before the invention of the telescope. These observations showed discrepancies within Ptolemy’s Geocentric theory of the universe.
- Johannes Kepler (1571 – 1630) studied data of motion of planets by Danish astronomer Tycho Brahe and put forward his laws of planetary motion. He also proposed that the motion of the planet around the sun is not in circular orbit but it is in an elliptical orbit with the Sun at one of the foci of the elliptical orbit.
- A Netherland’s spectacle maker Hans Lippershey assembled first reflecting telescope. Using this concept Galileo developed his own telescope. Using the telescope he observed skies and substantiated Copernicus’s Heliocentric theory of the universe. Further, he observed there are many stars in the universe and the Sun is one of them. He observed that only planets move around the sun. The moon moves around the Earth and Some planets like Jupiter has more than one moon.
- There was a curiosity among astronomers about the motion of planets around the sun.
- For the study of the motion of planets around the sun or satellites around the Earth, we consider gravitational attraction between the sun and the planet or earth and the satellite only. Here we ignore the disturbing effect of the gravitational forces due to other bodies. Another major assumption is that the centre of mass of the sun (earth) and the planet (satellite) system is at the centre of the earth.

**Keppler’s Laws:**

**First Law (Keppler’s Law of Elliptical Orbits – 1609):**

- Every planet moves around the sun in a closed elliptical orbit with the sun at one of its foci.
- The following figure shows the path of the planet around the Earth.

- Due to elliptical motion and the Sun at one of the foci, the distance of a planet from the sun changes continuously.
- The closest point on the orbit of the planet from the sun is B. This closest point on the orbit of the planet to the sun is called perihelion and this minimum distance is called perigee. Thus Perigee = SB = a – ae = a(1 – e). Where a is semi-major axis and e is the eccentricity of the ellipse.
- The farthest point on the orbit of the planet from the sun is A. This farthest point on the orbit of the planet to the sun is called aphelion and this maximum distance is called apogee. Thus Apogee = SA = a + ae = a(1 + e). Where a is semi-major axis and e is the eccentricity of the ellipse.

**Second Law (****Keppler’s ****Law of Equal Areas – 1609):**

- The radius vector drawn from the Sun to the planet sweeps out equal areas in equal time. i.e. aerial velocity of the radius vector is constant. Thus the Aerial velocity of the satellite is always constant. i.e. A/t = Constant.

- In this case, the planet covers unequal distances in equal time. Thus the planet has variable speed. When the planet is very close to the Sun, the maximum distance is covered in given time. Thus the planet has maximum velocity when it is at perihelion. Hence it has maximum kinetic energy at perihelion. When the planet is very far to the Sun, the minimum distance is covered in given time. Thus the planet has minimum velocity when it is at aphelion. Hence it has minimum kinetic energy at perihelion.

Point |
Perihelion | Aphelion |

Distance from sun |
Minimum | Maximum |

Distance called |
Perigee |
Apogee |

Speed of Planet | Maximum |
Minimum |

Kinetic energy of planet | Maximum |
Minimum |

Potential energy of planet | Minimum |
Maximum |

**Keppler’s Third Law (****Keppler’s Law of ****Period – 1618)**

- The square of the period of a satellite is directly proportional to the cube of the semimajor axis of its elliptical orbit.
- If T is the period of satellite and r is semimajor axis or the radius of the circular orbit of the satellite.

- A similar result is obtained for elliptical orbit and then the radius r of the circular orbit is replaced by semi major axis ‘a’.

**Note: **

- Keppler’s laws are empirical laws because they are based on observations and not on theory.
- Kepler’s laws are applicable whenever an inverse square law is involved. They are applicable to the solar system and artificial satellite systems.

**Proof of Keppler’s Law of Equal Areas:**

- Consider a planet moving in an elliptical orbit from point P
_{1}to point P_{2}, Let in a small interval of time Δt, it traces small area ΔA at the focus S. Let the angle traced by radius vector be Δθ at the Focus S.

The area of sector is given by

Where ω is angular velocity of the planet.

Now, the instantaneous angular momentum is given by

L = m r ω^{2}

Thus r ω^{2 }= L/m

Substituting in equation (1) we have

- As net torque acting on the planet is zero, the angular momentum is constant. Thus R.H.S. of equation (2) is constant.

Thus dA / dt = constant

Thus the areal velocity of the planet is constant. This proves Keppler’s second law.

**Proof of Newton’s Law of Gravitation Using Keppler’s Law:**

Let, m = Mass of planet

M = Mass of sun

R = Radius of orbit of the planet

ω= Angular velocity of the planet.

F = Centripetal force exerted on planet by the Sun

We know that

By Keppler’s law, we have T^{2} ∝ r^{3}

T^{2} = k r^{3 }

Substituting in equation(1)

The quantities in the bracket are constant.

- Thus force exerted by the sun on the planet is a) directly proportional to the mass ‘m’ of the planet. b) inversely proportional to the square of the distance ‘R’ between the planet and the sun. Now, the planet will exert equal and opposite force on the sun such that

- Now, the planet will exert equal and opposite force on the sun such that

- Thus force exerted by the planet on the Sun is a) directly proportional to mass ‘M’ of the Sun and b) inversely proportional to the square of the distance ‘R’ between the planet and the sun.

From (2) and (3) we have

This relation is known as Newton’s law of gravitation.

#### Example – 1:

- What would be the length of the year if the earth were at half its present distance from the sun?
**Solution:****Given:**r_{2}= 1/2 r_{1}, old period T_{1}= 365 days**To Find:**New period T_{2}=?

By Keppler’s law, we have T^{2} ∝ r^{3}

**Ans:** Thus length of the year would be 129 days

**Calculation Using Scientific Calculator:**

365 ÷ (1÷2)^(3÷2) =

**Calculation using Logarithmic Table:**

T_{2} = Antilog (- ½ log 8 + log 365) = Antilog (- ½ × 0.9031 + 2.5623)

T_{2} = Antilog (- 0.4516 + 2.5623) = Antilog (2.1107) = 1.290 × 10^{2} = 129.0

#### Example – 2:

- What would be the duration of the year if the distance of the Earth from the sun gets doubled?
**Solution:****Given:**r_{2}= 2 r_{1}, old period T_{1}= 365 days**To Find:**New period T_{2}=?

By Keppler’s law, we have T^{2} ∝ r^{3}

**Ans:** Thus length of the year would be 1032 days

**Calculation Using Scientific Calculator:**

(2)^(3÷2)×365 =

**Calculation using Logarithmic Table:**

T_{2} = Antilog (3/2 log 2 + log 365) = Antilog (3/2 × 0.3010 + 2.5623)

T_{2} = Antilog (0.4515 + 2.5623) = Antilog (3.0138) = 1.032 × 10^{3} = 1032

#### Example – 3:

- Calculate the period of revolution of the planet Jupiter around the Sun. The ratio of the radius of Jupiter’s orbit to that of earth’s orbit around the Sun is 5.2.
**Solution:****Given:**r_{J}: r_{e }= 5.2 , Time period of the Earth T_{1}= 1 year.**To Find:**Period of Jupiter T_{J}=?

By Keppler’s law, we have T^{2} ∝ r^{3}

**Ans:** The period of revolution of planet Jupiter is 11.86 years.

**Calculation Using Scientific Calculator:**

(5.2)^(3÷2) =

**Calculation using Logarithmic Table:**

T_{2} = Antilog (3/2 log 5.2) = Antilog (3/2 ×0.7160) = Antilog (1.0740) = 1.186 × 10^{1} = 11.86

#### Example – 4:

- A geostationary satellite is orbiting the earth at a height of 6R above its surface. What is the time period of a satellite orbiting at a height of 2.5 R above the earth’s surface? Where R is the radius of the Earth.
**Solution:****Given:**r_{1}= R + 6R = 7R, r_{2 }= R + 2.5 R = 3.5 R, Time period of geostationary satellite T_{1}= 24 hours.**To Find:**Period of second satellite T_{2}=?

By Keppler’s law, we have T^{2} ∝ r^{3}

**Ans:** The period of revolution of a satellite orbiting at a height of 2.5 R above the earth’s surface is 8.458 hr.

#### Example – 5:

- A satellite orbiting around the earth has a period of 8 hrs. If the distance of another satellite from the centre of the earth is four times that of above, what is its period?
**Solution:****Given:**Ratio of radii of orbits r_{1}= r, r_{2 }= 4r, Time period of first satellite T_{1}= 8 hours.**To Find:**Period of second satellite T_{2}=?

By Keppler’s law, we have T^{2} ∝ r^{3}

**Ans:** The period of the second satellite is 64 hours.

#### Example – 6:

- Calculate the period of revolution of a planet around the sun if the diameter of its orbit is 60 times that of Erath’s orbit around the Sun. Assume both orbits to be circular.
**Solution:****Given:**d_{P}= 60 d_{E}, r_{P}= 60 r_{E}, Time period of Earth T_{E}= 1 year.**To Find:**Period of the planet T_{P}=?

By Keppler’s law, we have T^{2} ∝ r^{3}

**Ans:** The period of the planet is 464.8 years

#### Example – 7:

- The planet Neptune travels around the sun with a period of 165 years. Show that the radius of its orbit is approximately thirty times that of the Earth.
**Solution:****Given:**Period of Neptune_{N}= 165 years, Time period of Earth T_{E}= 1 year.**To Show:**Radius of Neptune r_{N}=30 r_{E}

By Keppler’s law, we have T^{2} ∝ r^{3}

**Ans:** Thus the radius of its orbit is approximately thirty times that of the Earth.

#### Example – 8:

- The radius of earth’s orbit is 1.5 ×10
^{8}km and that of mars is 2.5 × 10^{11}m. in how many years the mars completes its one revolution. **Solution:****Given:**Radius of the orbit of the Earth r_{E}= 1.5 ×10^{8}km = 1.5 ×10^{11}m, Radius of the orbit of the Mars r_{M}= 2.5 ×10^{11}m, Time period of Earth T_{E}= 1 year.**To Find:**Time period of Mars T_{M}=?

By Keppler’s law, we have T^{2} ∝ r^{3}

T_{M} = 2.15 years

**Ans:** In 2.15 years Mars completes its one revolution.

#### Example – 9:

- The mean distance of the earth from the Sun is 1.496 ×10
^{8}km and its period of revolution around the Sun is 365.3 days. The time periods of revolutions of planets Venus and Mars are 224.7 days and 687.0 days respectively, where day means a terrestrial day. calculate the distances of Venus and Mars from the Sun. **Solution:****Given:**Radius of the orbit of the Earth r_{E}= 1.496 ×10^{8}km, Time period of Earth T_{E}= 365.3 days, Time period of Venus T_{E}= 224.7 days, Time period of Mars T_{M}= 687.0 days.**To Show:**Radius of the orbit of the Venus r_{V}= ?, Radius of the orbit of the Mars r_{M}= ?,

By Keppler’s law, we have T^{2} ∝ r^{3}

**Ans: **Distance of Venus from the sun is 1.08 ×10^{8} km and distance of Mars from the sun is 2.279 ×10^{8} km

#### Example – 10

- Suppose Earth’s orbital motion around the Sun is suddenly stopped. What time the Earth shall take to fall into the Sun.
**Solution:****Given:**Radius of the orbit of the Earth = r, Time period of Earth T = 365 days,**To Show:**Time taken by the Earth to fall into the Sun T_{2}= ?,- In this problem, we are assuming there is no effect of temperature on the Earth. If the Earth suddenly stops rotating around the sun and falls into the sun and let us assume it comes back to the point on the orbit. Thus it starts orbiting along a highly flattened ellipse with major axis = r. Thus semimajor axis = a = r
_{2}= r/2

By Keppler’s law, we have T^{2} ∝ r^{3}

This is the total period of revolution. It should take half the time period to fall into the Sun

Time required = 130/2 = 65 days

**Ans: **Earth shall take 65 days to fall into the Sun.

**Note:**Thus, in general, the time taken by planet to fall into the sun = Period x 0.1768 (This relation can be used for MCQ)

Science > Physics > Gravitation > You are Here |

Physics |
Chemistry |
Biology |
Mathematics |