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Johannes Kepler revealed three empirical laws by using the data on planetary motion, pertaining to the orbit of planets. These laws are contained here because of the huge importance they have regarding the motion of planets.
1. First law (Law of orbit): Every planet moves in an elliptical orbit, with the light at single focus of the ellipse. This is known as law of orbits. For simplification, the orbit can be supposed to be nearly circular because circle is special case of ellipse. Every planet has a definite orbit and dissimilar planets have different orbits.
2. Second law (Law of area): A line joining any planet to the sun (i.e. radius vector of the planet from sun) sweeps equivalent areas in equal interval of time i.e. the linear speed of the planet changes such that the areal velocity of planet is constant. The linear speed is maximum when the planet is nearest to sun, whereas it is minimum when the planet is farthest from the sun.
3. Third law (Law of period): The Square of the time periods of the planet is proportional to the cube of semi-major axis of ellipse. If T is time period of revolution of a planet around the sun and a is the semi-major axis of the orbit of ellipse, then
Radius of gyration may be explained as the distance from the axis at which, if the entire mass of the object were to be concentrated, the moment of inertia would be the same about
A 40 kilogram girl climbs a vertical distance of 5 meters in twenty seconds at a constant velocity. How much work has the girl done? Ans: 2000 joules or 1960 joules work is
short answer
(i) Conservation of charge number and mass number : In the given nuclear reaction Mass number (A) → Before the reaction After the reaction
my question is regarding value of pie whether its decimal value which has non-repeating nature goes to infinity or it ends at some place?
a report on electrostatic shielding
(a) A cast steel ring has a cross-sectional area of 600mm 2 and a radius of 25mm. Verify the mmf essential to establish a flux of 0.8mWb in the ring. Use the B-H curve for cast st
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Magnitude of K.E. in an orbit is equal to: (1)Half of the potential energy (2)Twice of the potential energy (3)One fourth of the potential energy (4 ) None
The beam below is supported at A and B. F1 and F2 in kN are given in P21 and P22 respectively. F3 in kN is given in P23 and the couple in kN.m is given in P24. Assume that F1, F
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