Chapter 13 Universal Gravitation

13.1 Newton’s Law of Universal Gravitation directly proportionalproduct of their massesinversely proportional distance between them Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the distance between them (13.1) (13.1) G  universal gravitational constant G =6.673 x N  m 2 / kg 2 (13.2) G  universal gravitational constant G = x N  m 2 / kg 2 (13.2)

Law of Gravitation, 2 inverse square law This is an example of an inverse square law The magnitude of the force varies as the inverse square of the separation of the particles The magnitude of the force varies as the inverse square of the separation of the particles The law can also be expressed in vector form(13.3) F 12 F 12 is the force exerted by particle 1 on particle 2 negative sign particle 2 is attracted toward particle 1 The negative sign in the vector form of the equation indicates that particle 2 is attracted toward particle 1 F 21 is the force exerted by particle 2 on particle 1

Law of Gravitation, 3 F 12 = – F 21 F 12 = – F 21 The forces form a Newton’s Third Law action-reaction pair Gravitation Gravitation is a field force that always exists between two particles, regardless of the medium between them forcedecreases distance increases The force decreases rapidly as distance increases inverse square law A consequence of the inverse square law

G vs. g Gg Always distinguish between G and g G universal gravitational constant G is the universal gravitational constant It is the same everywhere gacceleration due to gravity g is the acceleration due to gravity g = 9.80 m/s 2 g = 9.80 m/s 2 at the surface of the Earth g will vary by location g will vary by location

Gravitational Force Due to a Distribution of Mass (Earth) gravitational force finite-sizespherically symmetric mass distribution entire massconcentrated at the center The gravitational force exerted by a finite-size, spherically symmetric mass distribution on a particle outside the distribution is the same as if the entire mass of the distribution were concentrated at the center Earth: For the Earth:(13.4)

Newton’s Verification compared He compared the acceleration of the Moon in its orbit with the acceleration of an object falling near the Earth’s surface a M  (1/r M 2 ) & g  (1/R E 2 )  Using the Inverse Law Square for both cases: a M  (1/r M 2 ) & g  (1/R E 2 )  a M /g = (1/r M 2 ) / (1/R E 2 ) = R E 2 / r M 2 = 2.75x10 -4 a M /g = (1/r M 2 ) / (1/R E 2 ) = R E 2 / r M 2 = 2.75x10 -4  a M = 2.70x10 -3 m/s 2

Newton’s Verification, 2 (Centripetal Acceleration) calculated He calculated the centripetal acceleration of the Moon from its distance and period The Moon experiences a centripetal acceleration as it orbits the Earth  a M = 2.72x10 -3 m/s 2

Newton’s Verification Newton’s Verification, 3 (Assumption) Newton treated the Earth as if its mass were all concentrated at its center He found this very troubling Law of Universal Gravitation When he developed calculus, he showed this assumption was a natural consequence of the Law of Universal Gravitation inverse square nature of the law The high degree of agreement between the two techniques provided evidence of the inverse square nature of the law

13.3 Free-Fall Acceleration and The Gravitational Force (Finding g from G ) m mg The magnitude of the force acting on an object of mass m in freefall near the Earth’s surface is mg This can be set equal to the force of universal gravitation acting on the object (13.5) (13.5) 

g Above the Earth’s Surface h r R E + h If an object is some distance h above the Earth’s surface, r becomes R E + h(13.5) g decreasesincreasing altitude This shows that g decreases with increasing altitude r  weight zero As r , the weight of the object approaches zero

Example 13.1 “Weighing” Earth “Weighing” Earth! “Weighing” Earth! Determining mass of Earth!! M E M E = mass of Earth (unknown) R E R E = radius of Earth (known) R E = 6,370 km = 6.37 x 10 6 m Assumeperfect Assume the Earth is a uniform, perfect sphere. m m = mass of object on the surface of Earth (known) MEME RERE m

Example 13.1 “Weighing” Earth, final m The weight of m is: F g = mg The gravitational force on m is: F g = G(M E m)/R E 2 But these two are the same!!! But these two are the same!!! mg = G (mM E )/R E 2  mg = G (mM E )/R E 2  g = G M E /R E 2  g = G M E /R E 2  Solving for M E Solving for M E M E = g R E 2 /G = 5.98  kg MEME RERE m

Example 13.2 The Density of the Earth (Example 13.3 Text Book) average density Using the last result, find the average density of the Earth: From Example 13.1 From Example 13.1 mg = G (mM E )/R E 2 &g = G M E /R E 2  M E = g R E 2 /G  E = M E /V E & V E = (4/3)  R E 3  Since:  E = M E /V E & V E = (4/3)  R E 3 

F g 2r E Find F g acting on a 2000 kg spacecraft at 2r E R E = 6,380 km M E = 5.98  kg F G = G(mM E /r 2 ) r = r E At surface (r = r E ) F G = mg = G(mM E /r E 2 ) Example 13.3 Spacecraft at 2r E

r = 2r E At r = 2r E F G = G(mM E )/(2r E ) 2 F G = G(mM E )/(2r E ) 2  F G = G(mM E )/4r E 2 F G = G(mM E )/4r E 2  F G = ¼ G(mM E /r E 2 ) F G = ¼ G(mM E /r E 2 )  F G = ¼mg = 4900 N Example 13.3 Spacecraft at 2r E, final

Example 13.4 Geosynchronous Satellite geosynchronous satellite A geosynchronous satellite appears to remain over the same point on the Earth The gravitational force supplies a centripetal force m v h Gh R E M E A satellite of mass m is moving in a circular orbit around the Earth at a constant speed v and at an altitude h above the Earth’s surface. Find the speed of the satellite in terms of G, h, R E and M E

Example 13.4 Geosynchronous Satellite, final m Assume m is moving around the Earth in a circular orbit. Applying Newton’s 2 nd Law: ∑F = F g = m a R = mv 2 /r (1) F g =G(M E m)/r 2 (2) and: F g = G(M E m)/r 2 (2) (1) = (2) Taking (1) = (2) G(M E m)/r 2 = m v 2 /r G(M E m)/r 2 = m v 2 /r  v 2 = GM E /r(3) v 2 = GM E /r (3) r = R E +h Since r = R E +h  v 2 = GM E /(R E +h) v 2 = GM E /(R E +h)  v = [GM E /(R E +h)] ½

13.5 The Gravitational Field How it is possible for two objects to interact when they are not in contact with each other? How it is possible for two objects to interact when they are not in contact with each other? Newton himself could not answer that question force acting at a distance Even Newton’s contemporaries and successors found it difficult to accept the concept of a force acting at a distance Today: GRAVITAIONAL FIELD Today: We describe this interactions using the concept of a GRAVITAIONAL FIELD that exists at every point in space

The Gravitational Field, 2 (test particle)m gravitational fieldg F g = mg When a particle (test particle) of mass m is placed at a point where the gravitational field is g, the particle experiences a force: F g = mg fieldonparticle The field exerts a force on the particle gravitational field g The gravitational field g is defined as(13.9) gravitational field test particle test particle The gravitational field is the gravitational force experienced by a test particle placed at that point divided by the mass of the test particle

The Gravitational Field, 3 test particleg The presence of the test particle is not necessary for g to exist Earthg!!! Earth creates g!!! g ≡ F g /m F g = G (M E m)/r 2 At Earth’s surface: g ≡ F g /m But: F g = G (M E m)/r 2 (13.10) The unit vector points radially outward from the Earth. g Negative signs indicates that g points toward the center of the Earth. gravitational field“effect” would f The gravitational field describes the “effect” that any object has on the empty space around itself in terms of the force that would be present if a second object were somewhere in that space

The Gravitational Field, 4 gravitational field vectors the vicinity vary directionmagnitude!!! The gravitational field vectors in the vicinity of a uniform spherical mass such as the Earth vary in both direction and magnitude!!! gravitational field vectors direction acceleration The gravitational field vectors point in the direction of the acceleration a particle would experience if placed in that field magnitude freefall acceleration The magnitude of the field vector at any location is the magnitude of the freefall acceleration at that location

The Gravitational Field, final gravitational field vectorssmall regionnear uniform direction magnitude!!! The gravitational field vectors in a small region near the Earth’s surface are uniform in both direction and magnitude!!!

Examples to Read!!! Examples to Read!!! Example 13.2 Example 13.2 (page 395) Homework to be solved in Class!!! Homework to be solved in Class!!! Problem: 4 Problem: 4 Question: 6 Question: 6 Problem: 18 Problem: 18 Material for the Midterm

13.6 Gravitational Potential Energy U = mgy Valid only In Chapter 8: U = mgy (Particle-Earth). Valid only when particle is near the Earth’s surface. Inverse Square Law(F g  1/r 2 ) U Now with the introduction of the Inverse Square Law (F g  1/r 2 ) we must look for a general potential energy U function without restrictions.

Gravitational Potential Energy, cont fact We will accept as a fact that: gravitational force is conservative The gravitational force is conservative (Sec 8.3) gravitational force is a central force The gravitational force is a central force It is directed along a radial line toward the center r Its magnitude depends only on r A central force can be represented by

Grav. Potential Energy–Work AB A particle moves from A to B while acted on by a central force F The path is broken into a series of radial segments and arcs (F  v) r f r i Because the work done along the arcs is zero (F  v), the work done is independent of the path and depends only on r f and r i

Grav. Potential Energy–Work, 2 F The work done by F along any radial segment is 0 The work done by a force that is perpendicular to the displacement is 0 Total work Total work is Therefore, the work is independent of the path

Grav. Potential Energy–Work, final A B As a particle moves from A to B, its gravitational potential energy changes by (Recall Eqn 8.15)(13.11) general form This is the general form, we need to look at gravitational force specifically

Grav. Potential Energy for the Earth U i = 0r i → ∞ Taking U i = 0 at r i → ∞ we obtain the important result:(13.13)

Gravitational Potential Energy for the Earth, final Eqn r ≥ R E Eqn applies to Earth-Particle system, with r ≥ R E (above the Earth’s surface). Eqn r < R E Eqn is not valid for particle inside the Earth r < R E U U i = 0 r i → ∞ U is always negative (the force is attractive and we took U i = 0 at r i → ∞ An external agentyou lifting a book+ Work height U U r An external agent (you lifting a book) must do + Work to increase the height (separation between the Earth’s surface and the book). This work increases U  U becomes less negative as r increases.

Gravitational Potential Energy, General two particlesgravitational potential energy For any two particles, the gravitational potential energy function becomes(13.14) 1/r The gravitational potential energy between any two particles varies as 1/r force1/r 2 Remember the force varies as 1/r 2 The potential energy is negative because the force is attractive and we chose the potential energy to be zero at infinite separation

Systems with Three or More Particles total gravitational potential energy the sum The total gravitational potential energy of the system is the sum over all pairs of particles superposition principle Gravitational potential energy obeys the superposition principle U Each pair of particles contributes a term of U

Systems with Three or More Particles, final three particles Assuming three particles:(13.15) The absolute value of U total work neededseparate the particles The absolute value of U total represents the work needed to separate the particles by an infinite distance

Binding Energy binding energy The absolute value of the potential energy can be thought of as the binding energy larger excess kinetic energy If an external agent applies a force larger than the binding energy, the excess energy will be in the form of kinetic energy of the particles when they are at infinite separation

Example 13.5 The Change in Potential Energy (Example 13.6 Text Book) m  y A particle of mass m is displaced through a small vertical displacement  y near the Earth’s surface. Show that the general expression for the change in gravitational potential energy (Equation 13.12) reduces to the familiar relationship:  U = mg  y From Equation 13.12

Example 13.5 The Change in Potential Energy, cont r i r f Since r i and r f are to close to the Earth’s surface, then: r f – r i ≈  y r f r i ≈ R E 2 r f – r i ≈  y and r f r i ≈ R E 2 r Remember r is measured from center of the Earth  U = mg  y