Click to Chat
1800-2000-838
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Satellites and Planetary Motion It was Copernicus who, first of all, introduced the idea that the central body of our planetary system was Sun rather than Earth. Kepler later on confirmed by putting it on a solid mathematical back ground. Kepler announced two of his laws in 1609 and the third one 10 years later. The laws can be stated as below: (a) Kepler’s first law (law of elliptical orbit):- A planet moves round the sun in an elliptical orbit with sun situated at one of its foci. (b) Kepler’s second law (law of areal velocities):- A planet moves round the sun in such a way that its areal velocity is constant. (c) Kepler’s third law (law of time period):- A planet moves round the sun in such a way that the square of its period is proportional to the cube of semi major axis of its elliptical orbit. T^{2} ∝ r^{ 3} Here R is the radius of orbit. T^{2} = (4π^{2}/GM) r^{ 3} Earth and its Satellite:- Consider a satellite of mass m revolving in a circular orbit around the Earth, which is located at the centre of its orbit. If the satellite is at a height h above the Earth's surface, the radius of its orbit r = Re + h, where R_{e} is the radius of the Earth. The gravitational force between M_{e} & m provides the centripetal force necessary for circular motion, i.e., GM_{e}m/(R_{e}+h)^{2 }=^{ }mv^{2}/(R_{e}+h) Or v^{2} = GMe/(R_{e}+h) or v = √GMe/(R_{e}+h) Hence orbital velocity depends on the height of the satellite above Earth's surface. Time period T of the satellite is the time taken to complete one revolution. Therefore, T = 2Πr/v = 2Π(R_{e} + h)√(Re+h)/GMe or T^{2} = 4Π^{2}(R_{e}+h)^{3}/GMe where r = R_{e} + h If time period of a satellite is 24 hrs. Then, r = [GM_{e}T^{2}/4Π^{2}]^{1/3} = 42400 km and h = 36000 km. This gives the height of a satellite above the Earth's surface whose time period is same as that of Earth's. Such a satellite appears to be stationary when observed from the Earth's surface and is hence known as Geostationary satellite. For a satellite very close to the surface of Earth i.e. h << R_{e} then r ≈ R_{e} v_{orbital} = √GM_{e}/R_{e} = √gR_{e} Simulation for Kepler Laws of Motion:- Satellite in circular orbit:- For different velocities, the trajectory of the satellite would be different. Let us consider these cases. If v is the velocity given to a satellite and v_{O} represents the velocity of a circular orbit and v_{e} the escape velocity. i.e. v_{0} = √GMe/(Re+h) v_{e} = √2GMe/(Re+h) Where, h is the distance of the satellite from the surface of the Earth.
It was Copernicus who, first of all, introduced the idea that the central body of our planetary system was Sun rather than Earth. Kepler later on confirmed by putting it on a solid mathematical back ground. Kepler announced two of his laws in 1609 and the third one 10 years later. The laws can be stated as below:
(a) Kepler’s first law (law of elliptical orbit):- A planet moves round the sun in an elliptical orbit with sun situated at one of its foci.
(b) Kepler’s second law (law of areal velocities):- A planet moves round the sun in such a way that its areal velocity is constant.
(c) Kepler’s third law (law of time period):- A planet moves round the sun in such a way that the square of its period is proportional to the cube of semi major axis of its elliptical orbit.
T^{2} ∝ r^{ 3}
Here R is the radius of orbit.
T^{2} = (4π^{2}/GM) r^{ 3}
Consider a satellite of mass m revolving in a circular orbit around the Earth, which is located at the centre of its orbit. If the satellite is at a height h above the Earth's surface, the radius of its orbit r = Re + h, where R_{e} is the radius of the Earth. The gravitational force between M_{e} & m provides the centripetal force necessary for circular motion, i.e.,
GM_{e}m/(R_{e}+h)^{2 }=^{ }mv^{2}/(R_{e}+h)
Or v^{2} = GMe/(R_{e}+h) or v = √GMe/(R_{e}+h)
Hence orbital velocity depends on the height of the satellite above Earth's surface. Time period T of the satellite is the time taken to complete one revolution.
Therefore, T = 2Πr/v = 2Π(R_{e} + h)√(Re+h)/GMe
or T^{2} = 4Π^{2}(R_{e}+h)^{3}/GMe where r = R_{e} + h
If time period of a satellite is 24 hrs. Then,
r = [GM_{e}T^{2}/4Π^{2}]^{1/3} = 42400 km and h = 36000 km.
This gives the height of a satellite above the Earth's surface whose time period is same as that of Earth's. Such a satellite appears to be stationary when observed from the Earth's surface and is hence known as Geostationary satellite.
For a satellite very close to the surface of Earth i.e. h << R_{e} then
r ≈ R_{e}
v_{orbital} = √GM_{e}/R_{e} = √gR_{e}
For different velocities, the trajectory of the satellite would be different. Let us consider these cases.
If v is the velocity given to a satellite and v_{O} represents the velocity of a circular orbit and v_{e} the escape velocity.
i.e. v_{0} = √GMe/(Re+h)
v_{e} = √2GMe/(Re+h)
Where, h is the distance of the satellite from the surface of the Earth.
When, v < v_{0}, the satellite follows an elliptical path with center of the Earth as the further focus. In this case, if satellite is projected from near surface of the Earth, it will hit the Earth's surface without completing the orbit.
If v = v_{0}, obviously the satellite follows a circular orbit with center of Earth as the center of the orbit.
If v_{0} < v < v_{e}, then the satellite follows an elliptical orbit with center of the Earth as the nearer focus.
If v = v_{e}, the satellite escapes the gravitational field of the Earth along a parabolic trajectory.
If v > v_{e}, the satellite escapes the gravitational field of Earth along a hyperbolic trajectory
Question:-
What should be the energy required to shift a satellite orbiting around the Earth to infinity?
Answer:-
At infinity the Potential Energy of the satellite would be zero and if we want to supply minimum Energy then its kinetic energy would also be zero. Let us first find the total Energy of the satellite.
Total Energy = Kinetic Energy + Potential Energy
= 1/2 mv^{2} + (-GM_{e}m/r)
= 1/2 m (GM_{e}/r) - GM_{e}m/r
= -GM_{e}m/2r
Now, Binding Energy would be equal to - (Total Energy) as it is the energy needed to shift the satellite from its orbit to infinity.
So, the energy required = GMem/2r.
Here, r = R_{e} + h.
If we see a satellite from Earth, how long will it take for one revolution?
Let us consider a satellite in circular orbit with a time period T_{s}. The Earth also rotates with the time period T_{e} = 24 hrs. If an observer on Earth sees this satellite, the angular velocity of the satellite will be . Hence, the time period of revolution will seem different from T_{s} and will be observed as T_{SE}.
(i) If the satellite and Earth are revolving and rotating respectively in the same direction.
Hence, 2Π/T_{SE =} |2Π/T_{s} - 2Π/T_{e}|
=> T_{SE} = T_{s}T_{e}/|T_{e} - T_{s}|
(ii) If the satellite and Earth are revolving and rotating respectively in the opposite direction.
Hence,2Π/TSE = 2Π/T_{s }+_{ }2Π/T_{e}
or T_{SE} = T_{s}T_{e}/|T_{s} - T_{e}|.
One of the greatest ideas proposed in human history is the fact that the earth is a planet, among the other planets, that orbits the sun. The precise determination of these planetary orbits was carried out by Jhannes Kepler, using the data compiled by his teacher, the astronomer Tycho Brahe. Johannes Kelper discovered three empirical laws by using the data on planetary motion.
(a) Each planet moves in an elliptical orbit, with the sun at one foci of the ellipse.
(b) A line from the sun to a given planet sweeps out equal areas in equal intervals of time.
(c) The square of the periods of the planets are proportional to cube of their mean distance (or semi-major axis) from the sun.
These laws go by the name 'Kepler's laws of planetary motion'. It was in order to explain the origin of these laws, among other phenomena, that Newton proposed the theory of gravitation.
In our discussion, we are not going to derive the complete laws of planetary motion from Newton's law of gravitation. Since most of the planets actually revolve in near circular orbits, we're going to assume that the planets revolve in circular orbits.
Consider a planet of mass m rotating around the sun (mass M >> m) in a circular orbit of radius r with velocity v. Then, by applying Newton's law of gravitation and the second law of motion, we can write
Gravitational force = mass × centripetal acceleration
i.e. GMm/r^{2} = m(v^{2}/r) ... (1)
or, v^{2} = GMm/r ... (2)
As the moment of the gravitational force about S is zero, the angular momentum of the planet about the sun remains constant. This is the meaning of Kepler's 2nd law of motion, as will be shown later.
The time period of rotation, T, of the planet around the sun is given by,
T = 2Πr/v = -2Πr/√GM/r = 2Π/√GM r^{3/2} ... (3)
Squaring both sides,
T^{2} = (4Π^{2}/GM)r^{3} ... (4)
which is Kepler's 3^{rd} law of motion.
The constant of proportionality in the above equation depends only on the mass of the sun (M) but not on the mass of the planet.
Kepler's Laws are also valid for the motion satellites around the earth.
Consider a planet P that moves in an elliptical orbit around the sun, and let P and P' be the positions of the planet at time t and t + Δt (where Δt is a very small time interval). If the angular displacement of the planet is Δθ, then the are swept out by the line joining the planet and sun (SP) in time Δt is:
ΔA = area of the section SPP'
= 1/2 r^{2}.Δθ; where r = the length SP.
The area velocity, v_{A} = ΔA/Δt = 1/2r^{2}Δθ/Δt = 1/2r^{2}ω = constant .... (5)
In other words,
m × (2v_{A}) = constant as well (m = mass of the planet)
Areal velocity = dA/dt = L/2m ... (6)
This is the expression for the angular momentum of the planet,
L = Iω = mr^{2}ω
= mr^{2} (dθ/dt) perpendicular to the plane of its orbit.
The gravitational force,
= -GMm/r2 is centripetal, and the torque on the planet is zero,
So,
Hence, the angular momentum of the planet does not vary, i.e. the areal velocity of the planet remains constant. At its aphelion (farthest point from the sun, r is large), the planet moves slowly and at its perihelion (nearest point from the sun, r is small) the planet moves fastest.
The gravitational force between two particles are equal in magnitude but oppositely directed.
The gravitational force between two particles does not depend on the presence of the other bodies.
The gravitational force between two particles does not depend on the nature of the medium between the particles.
The force of gravitation is expressed in terms of the force between the masses of particles.
The value of gravitational force in the case of small bodies is very small but is appreciable in the case of massive bodies.
Problem 1:-
Calculate the mass of the Sun from the following data; distance between the Sun and the Earth = 1.49 × 10^{11} m, G = 6.67 × 10^{-11} SI units and one year = 365 days.
Solution:-
Force of attraction between the sun and the earth = Gm_{s}m_{E}/d_{SE}^{2}
Considering the orbit of the earth as nearly circular, the centripetal force acting on the earth is m_{E} d_{SE}ω^{2}.
=> m_{E} d_{SE} ω^{2} = Gm_{s}m_{E}/d_{SE}^{2}
m_{S} = d_{SE}^{2}.ω^{2}/G = 4Π^{2}d_{SE}^{2}/GT^{2}
=(4×(3.14)^{2}×(1.49×10^{11})^{2})/(6.67×10^{-11}×(365×24×60×60)^{2})
=1.32 × 10^{19} kg.
Problem 2:-
A Saturn year is 29.5 times the earth year. How far is Saturn from the sun (M) if the earth is 1.5 × 10^{8} km away from the sun?
It is given that
T_{S} = 29.5 T_{e}; R_{e} = 1.5 × 10^{11} m
Now, according to kepler's third law
T_{S}^{2}/T_{e}^{2} = R_{s}^{3}/R_{e}^{3}
R_{S}=R_{e}(T_{S}/T_{e})^{2/3}=1.5×10^{11} ((29.5T_{e})/T_{e})^{2⁄3}=1.43×10^{12} m =1.43×10^{9} km
Problem 3:-
A planet of mass m moves along an ellipse around the sun so that its maximum and minimum distances form the sun are equal to R and r, respectively. Find the angular momentum of this planet relative to the center of the sun.
According to Kepler's second law, the angular momentum of the planet is constant.
So, mv_{1}R = mv_{2}r or v_{1}R = v_{2}r
If the mass of the Sun is M, conserving total mechanical energy of the system at two given positions we have,
-GMm/R + 1/2 mv_{1}^{2} = -(G M m)/r + 1/2 mv_{2}^{2}
So, GM[1/R - 1/r] = v_{1}^{2}/2 + v_{2}^{2}/2 or GM[(r-R)/Rr] = v_{1}^{2}/2 + v_{2}^{2}R^{2}/2r^{2}
Thus, v_{1}^{2}= (2GM(R-r) r^{2})/Rr(R^{2}-r^{2} ) = 2GMr/R(R+r)
Now, angular momentum = mv_{1}R = m√2GMR/(R+r)
Question 1:-
Gravitational forces are:
(a) always attractive
(b) always repulsive
(c) sometimes attractive and sometimes repulsive
(d) none
Question 2:-
The atmosphere is held to the earth by:
(a) winds (b) gravity (c) clouds (d) the rotation of the earth
Question 3:-
Weight of the body is maximum:
(a) on the poles
(b) on the equator
(c) at a place in between pole and equator
Question 4:-
A body mass m is taken to the bottom of a deep mine. Then :
(a) its mass increases
(b) its mass decreases
(c) its weight increases
(d) its weight decreases
Question 5:-
If V_{e} and V_{P} denote the escape velocities from the earth and from another planet having twice the radius and the same mean density as the earth. Then:
(a) V_{e} = V_{P} (b) V_{e} = V_{P}/2 (c) V_{e} = 2V_{P} (d) V_{e} = V_{P}/4
a
b
d
You might like to refer Gravitational Potential Energy.
For getting an idea of the type of questions asked, refer the Previous Year Question Papers.
Click here to refer the most Useful Books of Physics.
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
Gravitational Potential Energy When two or more...
Solved Examples on Gravitation Download IIT JEE...
Newton’s Law of Gravitation:- Statement:-...
Gravitational Field and Intensity The space around...