1 The law of gravitation can be written in a vector notation (9.1) Although this law applies strictly to particles, it can be also used to real bodies with sizes small in comparison to the distance between them. It can be also applied to massive objects of a spherical symmetry. 9. Gravitation 9.1. Newton’s law of gravitation Two particles attract each other with a gravitational force proportional to the product of their masses m 1 and m 2 – masses of particles r – separation between particles G = 6.67 · Nm 2 /kg 2, the gravitational constant

Gravitational acceleration of the Earth The Earth attracts mass m located at a distance r from its center with a force (9.2) The gravitational force F causes falling of a mass m toward the Earth’s center with acceleration (9.3) Gravitational acceleration varies with the altitude h = r – R Z above surface of the Earth. From eq.(9.3) one obtains:h (km)a g (m/s 2 ) (comm. satellite) We must however take into account that the Earth is not a sphere, its mass is not uniformly distributed and it rotates about an axis. Therefore the assumption that acceleration g = 9.8 m/s 2 at any place on Earth’s surface is only a simplification. In general the weight Q = mg is a bit less (by 0.034m/s 2 at the equator) than the gravitational force given by eq.(9.2).

Motion of planets and satellites In that motion the Newton’s laws of dynamics and the law of gravitation hold. In the figure below two objects with comparable masses rotate around the common center of mass C. They have the same angle velocities ω. From the Newton’s third law the centripetal forces acting on each mass are equal and equal to the gravitational force (9.4) (9.5) For the system star – planet the mass of a star is much greater than that of a planet (M>>m), hence R<<r. In this case one obtains from eq.(9.5) (9.6) In general the planets move in elliptical orbits with a star (e.g. the Sun) at one focus (first Kepler’s law) and in this case r in eq.(9.5) is replaced with a semimajor axis a.

4 Motion of planets and satellites, cont. A relation between T 2 and r 3 obtained from (9.6) is also known as the third Kepler’s law. One can determin from it the mass of a star if we know T and r for the orbitting planet. Eq.(9.6) can be also used to analyse the motion of satellites orbitting the Earth (in this case we substitute M = M Z ). Example Using eq.(9.6) calculate the radius of a geosynchronous orbit, at which the satellite seems to be immobile for the Earth’s observer. In this case a plane of the satellite’s orbit must coincide with a plane of the equator. (9.7) From (9.7) we calculate r (9.8) Substituting the required data one obtains for r ca km.

Gravitational potential energy Gravitational force is a typical conservative force. In Section 5.4 we introduced the expression for the potential energy at point B given by vector r in reference to the potential energy U A at point A Assuming that U A = 0 for r = ∞ one obtains for the potential energy of a particle in the field of conservative forces (9.9) If the source of a conservative force F C is the Earth, one obtains from (9.9) (9.10) Eq.(9.10) is the potential energy of a mass m in the field of mass M (strictly speaking it is the potential energy of a system of two masses).

6 Gravitational potential energy, cont. Escape speed What is the necessary speed of firing upward from the Earth surface a projectile which will not come back to Earth (escape speed from Earth). From eq.(9.10) it follows that in infinity the potential energy of a projectile is zero. In infinity the projectile stops, so the kinetic energy is also zero. Its total mechanical energy in infinity is therefore zero. From the principle of conservation of mechanical energy it follows that the sum of potential and kinetic energies at the Earth’s surface is equal to the sum of energies in infinity, i.e. zero. (9.11) This gives (9.12) From (9.12) it is possible to calculate the escape speed from any astronomical object. For the neutron star one obtains the speed of order 10 5 km/s.

7 Energy in orbital motion A satellite of mass m orbits Earth of mass M in a radial orbit. Because M >> m, we can assign the kinetic and potential energies of the system to the satellite alone. The potential energy is given by eq.(9.10) (9.13) The kinetic energy can be found from the condition that the gravitational force is a centripetal force acting on the satellite (9.14) The mechanical energy is then equal (9.15) The total energy is negativ what means that the orbit is closed. For the motion in an elliptical orbit, we change r in eq.(9.15) for the semimajor axis a. During motion in an elliptical orbit the kinetic and potential energies vary periodically, but the total mechanical energy is always constant.