I.B. Gravity of a Solid Sphere

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Presentation transcript:

I.B. Gravity of a Solid Sphere Graph of |Fg| vs. r for a solid sphere of uniform density r: |Fg(inside)| = Gm(r)m/ r2 = G{(4p/3)rr3}/r2 ~ r. |Fg(outside)| = GMEm/ r2 = ~ r-2. |Fg| R ~ 1/r2 ~ r r (I.B.1) (I.B.2)

I.B. Gravity of a Solid Sphere 2. Graph of |Fg| vs. r for a solid sphere with M(r) = Mtotr2 |Fg(inside)| = Gm(r)m/ r2 = G{Mtotr2}/r2 ~ constant. |Fg(outside)| = GMEm/ r2 = ~ r-2. r ~ r-1. |Fg| R ~ 1/r2 ~ r r (I.B.3) (I.B.4)

I.B. Gravity of a Solid Sphere Inside a sphere, form of gravitational force depends on the distribution of matter--the density. M(r) = ∫rdV. (I.B.5) 4. Outside a sphere, form of gravitational force depends on the inverse square law. 5. Far enough away from any finte matter distribution, |Fg|~ r-2. At Earth’s surface: Fg = GmEm/RE2 This is just the weight of mass m at the surface: W = Fg = GmEm/RE2 = mg, where g = GmE/RE2. (I.B.6)

I.C. Gravitational Potential Energy For uniform gravity, U = mgy (e.g., near Earth’s surface) We need a more general expression for U based on Eqn. I.A.1 As before, we define potential energy in terms of the work done by the force associated with the potential energy: This leads to the following definition of gravitational potential energy: U = –GmEm/r (I.C.1)

U increases (becomes less negative) with increasing r I.C. Gravitational Potential Energy 5. For distances greater than the Earth’s radius: This definition of U means that U = 0 when r =  What about a hollow sphere? U RE r RE U increases (becomes less negative) with increasing r –GmEm / RE

I.D. Escape Speed 1. Consider a projectile fired from a cannon If gravitational force is the only force that does work, then mechanical energy is conserved: K1 + U1 = K2 + U2 For rocket to just barely escape to large values of r (say r = ), K2 = 0 At r = , U2 = 0 as well Therefore K1 + U1 = 0, where K1 and U1 are measured at the launch point (ground): 1/2mv2 + (–GmEm/ RE) = 0  vesc = [(2GmE)/ RE]1/2 = (2gRE)1/2 (I.D.1) = 11.2 km/s ~ 24,000 mph

I.E. Satellite Motion c) From Newton’s 2nd Law: GmEm / r2 = marad = mv2 / r, so vc = (GmE /r)1/2 = vesc/√2 (I.E.1) “Circular Orbit Speed” In terms of the period T of the orbit: (II.E.2)

I.E. Satellite Motion Example: What is the orbital period for “Low Earth Orbit?” For LEO, assume that r ~ RE (good assumption since r is typically RE + 300-500 km) and circular orbit: T ~ 2pRE3/2/√(GME) ~ 90 minutes. note: vc(LEO) ~ 8 km/s = 17,000 mph

I.E. Satellite Motion Example: What is the orbital radius for a geosynchronus orbit? Ts/TLEO = 24 hrs/1.5 hrs = r3/2/RE 3/2; rs/RE = 162/3 = 6.3, or rs ~ 6RE ~ 40,000 km. vs /vc(LEO)= (RE/rS)1/2 = 0.4, or vs = 0.44vc(LEO) ~ 3 km/s