2. Topic 2.4 Extended B – Gravitation inside the earth N EWTON’S S HELL T HEOREM  A uniform spherical shell of matter exerts no net gravitational force on a particle located inside it. m M F mM = 0 For a sphere, the net force from opposite conic sections exactly counter-balance one another… m M F mM = 0 More mass Farther away Less mass Closer No matter where inside the sphere the particle is located. Force on particle inside spherical shell F mM = 0

3. Topic 2.4 Extended B – Gravitation inside the earth N EWTON’S S HELL T HEOREM  A uniform spherical shell of matter exerts a net force on a particle located outside it as if all the mass of the shell were located at its center. m M r Note: Proof of Newton’s shell theorem required the use of calculus. This is the reason Newton invented integral calculus! Force on particle outside spherical shell F mM = GmM r 2

4. Topic 2.4 Extended B – Gravitation inside the earth  Even though the earth is not homogeneous, we can use Newton’s shell theorem to prove that we were justified in treating the earth as a point mass located at its center in all of our calculations inner core M i outer core M o mantle M m crust M c  For a point mass m located a distance r from the center of the earth we have F mM = GmM c r 2 + GmM m r 2 + GmM o r 2 + GmM i r 2  so that F mM = Gm(M c +M m +M o +M i ) r 2 F mM = GmM E r 2

5. Topic 2.4 Extended B – Gravitation inside the earth  Just for fun, suppose we drill a hole through the earth with the intention of using a freefall capsule for travel from one side of the earth to the other:  Because of Newton’s shell theorem, only the mass M’ inside the current radius of the capsule will contribute to the gravitational force acting on the mass: r M’ F = - GmM’ r 2  where M’ = ρV’ 4πr334πr33 M’ = ρ  so that F = - r 4πρGm 3 F = - Gm4πρr 3 3r 2 Gravitational force on mass inside the earth Note: ρ = M E /V E = 3M E /4πR E 3 = 5523 kg/m 3.

6. Topic 2.4 Extended B – Gravitation inside the earth  Just for fun, suppose we drill a hole through the earth with the intention of using a freefall capsule for travel from one side of the earth to the other:  Then r M’ F = - r 4πρGm 3 r” = - r 4πρG34πρG3 ma = - r 4πρGm 3 r” = -ω 2 r  where 4πρG34πρG3 ω = Angular frequency for capsule through the earth ρ = 5523 kg/m 3

7. Topic 2.4 Extended B – Gravitation inside the earth  Just for fun, suppose we drill a hole through the earth with the intention of using a freefall capsule for travel from one side of the earth to the other:  Then r M’ 4πρG34πρG3 ω = 4π(5523)6.67×10 -11 3 ω = ω = 1.2422 × 10 -3 rad/s  and 2πω2πω T = = 5058 s = 84 min Note: It would take only 42 minutes to reach the other side of the earth.  Since v max = r m ω, v max = R E ω = (6.37 × 10 6 )(1.2422 × 10 -3 ) = 7913 m/s !