Created by the Best Teachers and used by over 51,00,000 students. NCERT Exemplar Class 11 Physics Chapter 7 Gravitation Multiple Choice Questions Single Correct Answer Type Q1. NCERT Solutions for Class 11 Physics Physics Chapter 6 Work Energy and power includes all the important topics with detailed explanation that aims to help students to understand the concepts better. m= mass of the satellite, v=velocity of the satellite. Choose the correct alternative: The energy required to launch an orbiting satellite out of the earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as of the satellite) out of the earth’s influence. Here we have given NCERT Exemplar Class 11 Physics Chapter 7 Gravitation. The earth will then be at one of the foci of this ellipse. NCERT Exemplar Class 11 Physics Chapter 7 Gravitation are part of NCERT Exemplar Class 11 Physics. During the radial portion, the force F is opposite to the direction we are traveling along with distance dr. Ans: Kinetic energy is always positive because mass cannot be negative and the square of velocity gives a non-negative number. (a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy. From the derivation of Kepler’s third law, =(4x3.14x3.14x(3.84)3x1024)/6.67x10-11x(27.3x24x60x60)2. Let's think a bit about the total energy of orbiting objects. Thus, the total energy of an orbiting satellite at infinity is equal to the negative of its kinetic energy. It requires relatively lesser energy to move out of the influence of the Earth’s gravitational field than a stationary object on the Earth’s surface that initially contains no energy. This energy is a negative value of the total energy of the satellite. If h is the height of the satellite above the Earth’s surface and R is the radius of the Earth, then the radius of the orbit of satellite is r = R + h. If m is the mass of the satellite, its potential energy is, E P = -GMm/r = -GMm/(R+h) As the Earth-satellite system is a bound system, the total energy of the satellite is negative. The total energy of the satellite is calculated as the sum of the kinetic energy and the potential energy, given by. the kinetic energy of the satellite in a circular orbit with speed v is `"K.E" = 1/2"mv"^2` `="GmM"_E/(2("R"_E+"h"))` Considering gravitational potential energy at infinity to be zero, the potential energy at … Along the arc, F is perpendicular to dr, so F.dr = 0. However, the change in kinetic energy ∆K can be negative. A satellite revolving in a circular orbit round the Earth possesses both potential energy and kinetic energy. That gives us K orbit = 2.98 × 10 11 − 3.32 × 10 10 = 2.65 × 10 11 J K orbit = 2.98 × 10 11 − 3.32 × 10 10 = 2.65 × 10 11 J. Pro Lite, Vedantu (b) An orbiting satellite acquires a certain amount of energy that enables it to revolve around the Earth. Problem:- Express the constant k of in Eq. It is bound to the earth just like the earth is bound to the sun. Both methods yield almost the same answer, the difference between them being less than 1%. Energy of an orbiting satellite; Geostationary Satellite; ... Class 11 Physics Gravitation: Earth Satellites: Earth Satellites. + P.E. To make this TE as zero, we need to give an additional energy of GMm/2r to the satellite, i.e: We know that if the separation between the two bodies is infinite, then the potential of the system is considered zero. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. The two forces are acting on it are gravitational force, F, and a centripetal force due to its velocity, F, Gravitational Potential Energy of Satellite, Potential Energy of Charges in an Electric Field, Satellite Communication Active and Passive Satellite, Vedantu EduRev, the Education Revolution! Mass of the Satellite = 200 Kg; Mass of the Earth = 6.0 ×1024 Kg; Radius of the Earth = 6.4 ×106 M Concept: Energy of an Orbiting Satellite. Energy of an orbiting satellite: m = mass of the satellite, v = velocity of the satellite. Ans: In a circular motion, the satellite remains at the same distance above the earth’s surface; its radius of the orbit is fixed, and its speed remains constant. ’ t have GPS, nearest Starbucks, etc ’ s third law, = ( (... Function to compute the kinetic energy of the satellite available for now to bookmark for Online... Any point on its surface = kinetic energy to semi-major axis for earth... 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