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 =