Astroparticle Physics

378 17 SolutionsThis results inM planet ≈ 5.9 × 10 20 kg .The result of this estimate depends very much on earthlike properties and the assumptionthat other effects, like rotational velocity (→ centrifugal force) are negligible.4. a) The long-range forces in three dimensions are the gravitational force and the electromagneticforce, the latter of which reduces in the non-relativistic (static) limit tothe Coulomb force. The force is proportional to the masses (Newton’s gravitationlaw) or to the charges (Coulomb’s law), respectively. Both forces scale with 1/r 2 ,where r is the distance between the two bodies. The surface of a three-dimensionalsphere is 4πr 2 , and the forces are, roughly speaking, proportional to the solid angleunder which the bodies see each other. The surface of an n-dimensional sphere is(already from dimensional considerations) given bys n (r) = s n r n−1 , i.e., g n = n − 1 ,where the constant s n is the surface of the corresponding unit sphere. Isotropic longrangetwo-body forces in n dimensions should therefore scale asF(r) ∼ 1s n (r) ∼ 1r n−1 .b) The potential of a radial force is given by (F · r < 0 for attractive forces, n>2)∫ r∫ (rV(r)= F(r ′ ) dr ′ dr ′ 1 1∼ =−r 0 r 0r ′n−1 n − 2 r n−2 − 1)r n−20∼− 1 + const .rn−2 A particle of mass m, at distance r from a center, and having velocity v on a circularorbit has an angular momentum of L = rp = mvr. (In general, the velocitycomponent perpendicular to r is considered.) The centrifugal force is then given byF c (r) = mv2r= L2mr 3 ,where the last expression is also valid for general motion and L is a constant forcentral forces. The corresponding centrifugal potential therefore reads∫ rV c (r) =− F c (r ′ ) dr ′ =L2r 02mr 2 + const .The effective potential is the sum of V and V c ,V eff (r) = V(r)+ V c (r) =− C nr n−2 +L22mr 2 + const .Circular orbits take place for a vanishing effective radial force,