1 / 9

Graph of | F g | vs. r for a solid sphere of uniform density r :

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

perry-emerson
Download Presentation

Graph of | F g | vs. r for a solid sphere of uniform density r :

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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)

  2. I.B. Gravity of a Solid Sphere r ~ r-1. 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. |Fg| R ~ 1/r2 ~ r r (I.B.3) (I.B.4)

  3. 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)

  4. 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)

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

  6. 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

  7. 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)

  8. 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

  9. 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

More Related