CALC A shaft is drilled from the surface to the center of the earth (see Fig. 13.24). As in Example 13.10 (Section 13.6), make the unrealistic assumption that the density of the earth is uniform. With this approximation, the gravitational force on an object with mass m, that is inside the earth at a distance r from the center, has magnitude (as shown in Example 13.10) and points toward the center of the earth. (a) Derive an expression for the gravitational potential energy U(r) of the object–earth system as a function of the object’s distance from the center of the earth. Take the potential energy to be zero when the object is at the center of the earth. (b) If an object is released in the shaft at the earth’s surface, what speed will it have when it reaches the center of the earth?
Solution 85P Step 1 Gm m r The gravitational force at a point inside the earth, F = E g RE 3 Where, G - universal gravitational constant mE mass of earth m - mass of the object r - Distance of the object from the centre of the earth and r < R E RE Radius of the earth We know that, the gravitational force, F = - dU/dr g Where, U - Gravitational potential Step 2 a) We can write, dU = - F . g r Or, U = F g dr 0 r That is, U = GmEm rdr RE 3 0 r Gm E Taking the constants outside, U = R 3 r dr E 0 Integrating the equation, we get, Gm m 2 U = 3 [ 2 0 to r RE Gm m r 2 U = 1 E 2 3 R E But, we know that, the gravitational potential at the surface of the earth is, U = -Gm m/R 2 E E So, the potential should be continuous at this limit. Therefore, we can write,Ef r = R , Gm m r 2 Gm m 1 E 3 + C = E 2 RE RE Where “C” is a constant. Eut r = R Gm m R 2 Gm m Then, 1 E 3 E + C = E 2 RE RE Or, 1Gm E + C = Gm E 2 RE RE Gm m Gm m That iC = E 1 E RE 2 RE Gm m C = 3 E 2 RE Therefore, the gravitational potential at a point “r” inside the earth is, 3 Gm E 1 Gm E r 2 U = 2 R + 2 3 E RE Rearranging, we get, 1 Gm E r2 U = 2 RE [3 R 2 ] E