GRAVITATION

traverses a distance π ( R E + h ) with speed V . Its time period T therefore is / ( ( ( . ) on substitution of value of V from Eq. ( .

). Squaring both sides of Eq. ( . ), we get T = k ( R E + h) (where k = π / GM E ) ( .

) which is Kepler’s law of periods, as applied to motion of satellites around the earth. For a satellite very close to the surface of earth h can be neglected in comparison to R E in Eq. ( . ).

Hence, for such satellites, T is T o , where ( . ) If we substitute the numerical values g ≃ . m s - and R E = km., we get . .

s Which is approximately minutes. Example . The planet Mars has two moons, phobos and delmos. (i) phobos has a period hours, minutes and an orbital radius of .

× km. Calculate the mass of mars. (ii) Assume that earth and mars move in circular orbits around the sun, with the martian orbit being . times the orbital radius of the earth.

What is the length of the martian year in days ? Answer (i) We employ Eq. ( . ) with the sun’s mass replaced by the martian mass M m = M m .

- m = = . × kg. (ii) Once again Kepler’s third law comes to our aid, MS ES where R MS is the mars -sun distance and R ES is the earth-sun distance. ∴ T M = ( .

) / × = days We note that the orbits of all planets except Mercury and Mars are very close to being circular. For example, the ratio of the semi- minor to semi-major axis for our Earth is, b/a = .99986. Example . Weighing the Earth : You are given the following data: g = .

ms – , R E = . × m, the distance to the moon R = . × m and the time period of the moon’s revolution is . days.

Obtain the mass of the Earth M E in two different ways. Answer From Eq. ( . ) we have .

The moon is a satellite of the Earth. From the derivation of Kepler’s third law [see Eq. ( . )] M E .

- . kg Both methods yield almost the same answer, the difference between them being less than %. Example . Express the constant k of Eq.

( . ) in days and kilometres. Given k = – s m – . The moon is at a distance of .

× km from the earth. Obtain its time-period of revolution in days. Answer Given k = – s m – km                 = . × – d km – Using Eq.

( . ) and the given value of k, the time period of the moon is T = ( . × - )( . × ) T = .

d Note that Eq. ( . ) also holds for elliptical orbits if we replace ( R E + h ) by the semi-major axis of the ellipse. The earth will then be at one of the foci of this ellipse.

. ENERGY OF AN ORBITING SATELLITE Using Eq. ( . ), the kinetic energy of the satellite in a circular orbit with speed v is K E m v ( Gm M , ( .

) Considering gravitational potential energy at infinity to be zero, the potential energy at distance (R e +h) from the centre of the earth is G m M P E ( . ) The K.E is positive whereas the P.E is negative. However, in magnitude the K.E. is half 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