📖 generic · CBSE Class 11 English medium · PHYSICS · Page 11question

traverses a distance 2 π ( R E + h ) with speed V . Its · Part 3

Chapter 7: GRAVITATION · PHYSICS

the P.E, so that the total E is ( G m M K E P E ( . ) The total energy of an circularly orbiting satellite is thus negative, with the potential energy being negative but twice is magnitude of the positive kinetic energy. When the orbit of a satellite becomes elliptic, both the K.E. and P.E.

vary from point to point. The total energy which remains constant is negative as in the circular orbit case. This is what we expect, since as we have discussed before if the total energy is positive or zero, the object escapes to infinity. Satellites are always at finite distance from the earth and hence their energies cannot be positive or zero.

SUMMARY . Newton’s law of universal gravitation states that the gravitational force of attraction between any two particles of masses m and m separated by a distance r has the magnitude m m r where G is the universal gravitational constant, which has the value . × – N m kg – . .

If we have to find the resultant gravitational force acting on the particle m due to a number of masses M , M , …. M n etc. we use the principle of superposition. Let F , F , ….

F n be the individual forces due to M , M , …. M n , each given by the law of gravitation. From the principle of superposition each force acts independently and uninfluenced by the other bodies. The resultant force F R is then found by vector addition F R = F + F + ……+ F n = F i n

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