Chapter 12Gravity

DEHS 2011-12Physics 1

Newton’s Law of Universal Gravity• The force of gravity between any two point objects

of mass m1 and m2 is attractive and of magnitude

• Here, r is the distance between the masses

• G is called the universal gravitation constant and it is equal to €

F =Gm1m2

r2

€

G = 6.67 ×10−11 N ⋅m2 kg2

Inverse Square Dependence & Superposition

• Newton’s Law of Universal Gravitation (LUG) is an example of an inverse square dependence– The strength of gravity falls off

rapidly with distance, but it never becomes zero

– Gravity has an infinite range• The net gravitational force like

other field forces can be found using the principle of superposition, it is the vector sum of each other the forces individually

Example 12-1An astronaut is a distance d from his spaceship when he

experiences a gravitational pull of 80 N from his ship. He drifts out to a distance of 4d. Now what gravitational force does the ship exert on the astronaut?

Example 12-2The picture shows an arrangement of three particles, particle 1 of mass

m1 = 6.0 kg, particles 2 and 3 of mass m2 = m3 = 4.0 kg, and a distance a = 2.0 cm. What is the net gravitational force on particle 1 due to the other particles?

Example 12-3The picture shows an arrangement of five particles, with masses

m1 = 8.0 kg, m2 = m3 = m4 = m5 = 2.0 kg, and with a = 2.0 cm and θ = 30°

Gravity due to a SphereThe net force exerted

by a sphere whose mass is SPHERICALLY distributed (does not have to by uniformly distributed) throughout an object is the same as if all of the sphere’s mass were concentrated at its center

The Cavendish Experiment – “Weighing the Earth”

“Weighing the Earth” and consequences

• Cavendish didn’t actually weigh the Earth, he accurately measured the value for G

• Prior to his experiment, the quantities g and RE were known from direct measurement

Calculate the mass of Earth:

“Weighing the Earth” and consequences

• Calculate the volume of the Earth:

• Calculate the average density of the Earth:

Typical rocks found near the surface of the Earth, have a density of about 3.00 g/cm3. What can you conclude?

Acceleration due to gravity• Acceleration due to gravity (g) varies by location• g is determined by the distance from the center to

the planet it is near and the mass of the planet

€

g =GM

r2

€

r = R + h

g as a function of h

Example 12-4A hypothetical planet has a mass of 2.5 times that of Earth,

but the same surface gravity as Earth. What is this planet’s radius (in Earth radii)?

Example 12-5You are explaining to friends why astronauts feel weightless orbiting in

the space shuttle, and they respond that they thought gravity was just a lot weaker up there. Convince them by calculating how much weaker gravity is 300 km above Earth’s surface.

Example 12-6Certain neutron stars (extremely dense stars) are believed to be rotating at

about 1 rev/s. If such a star has a radius of 20 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation?

Some important Astronomy Terms• Astronomical Unit (AU): is the mean distance from

the Earth to the Sun (1 AU= 1.5×1011 m)• Periapsis: The point of closest approach of an object

to the body being orbited– perihelion: closest distance from the Sun– perigee: closest distance from the Earth

• Apoapsis: The point of furthest excision of an object from the body being orbited– aphelion: furthest distance from the Sun– apogee: furthest distance from the Earth

Kepler’s 1st Law of Orbit – Law of Elliptical OrbitsPlanets follow elliptical orbits, with the Sun at one

focus of the ellipse• The semi-major axis (a) is also equal to the average

distance from a focus to points on the ellipse

Orbital Distance at Apoapsis and Periapsis

• The eccentricity (e) is defined so that ea is the distance from the center of the ellipse to the either focus

• In circular orbits the two foci merge to one focus and e = 0

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Ra = a 1+ e( )

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Rp = a 1− e( )

Orbital Shape depends on eccentricity!• All orbits take the shape of a conic

section, and the shape depends on the eccentricity

• Closed orbits includes circular orbits (e = 0) and elliptical orbits (0 < e < 1)

• Open trajectories include parabolas and hyperbolas where an object approach the body and will never return (an escape trajectory)– Parabolic trajectory (e = 1), object will

reach infinity with zero KE (E = 0)– Hyperbolic trajectory (e > 1) object will

reach infinity with excess KE (E > 0)

Kepler’s 2nd Law of Orbit – Law of Equal Areas

As a planet moves in its orbit, it sweeps out an equal amount of area in an equal amount of time

• This means that they move faster when it is closer to the Sun

• This is a consequence of the conservation of angular momentum

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v1r1 = v2r2

Example 12-7The Earth’s orbit around the Sun is in the shape of an ellipse. Its aphelion is

152,000,000 km. Its perihelion, is 147,000,000 km. The Earth’s speed is 30,300 at perihelion. What is the Earth’s orbital speed at aphelion?

Kepler’s 3rd Law of Orbit – Law of Harmonics

The period, T, of a planet is proportional to its mean distance from the Sun raised to the 3/2 power

€

T = constant( )a3 / 2

€

T 2

a3=

4π 2

GM

⎛

⎝ ⎜

⎞

⎠ ⎟

or

Derive Kepler’s 3rd Law

Example 12-8Comet Halley orbits the Sun with a with a period of 76 years, and in

1986, had a distance of closest approach to the Sun, its perihelion distance RP, of 8.9×1010 m. (a) What is Halley’s average distance from the Sun? (b) What is the eccentricity of comet Halley? (c) What is the comet’s farthest distance from the Sun, its aphelion distance?

Example 12-9We SEE nothing at the exact center of our galaxy, the Milky Way. However, we

know that something has to be there, because all of the stars in our galaxy orbit this point. We can observe a star, called S2 as it moves around this mysterious object called Sagittarius A*. S2 orbits Sagittarius A* with a period of 15.2 years and with a semimajor axis of 5.5 light days (= 1.42×1014 m). What is the mass of Sagittarius A*? What is Sagittarius A*?

Kepler’s 3rd Law (alternate form)• Because the ratio T2/a3 is a constant for any objects

in orbit around the same mass, we can say

• When comparing to Earth, remember a = 1 AU and T = 1 yr

€

T12

a13

=T2

2

a23

or

€

T1

T2

⎛

⎝ ⎜

⎞

⎠ ⎟

2

=a1

a2

⎛

⎝ ⎜

⎞

⎠ ⎟

3

Example 12-10(a) Uranus has a semi-major axis of 19.23 AU. Calculate Uranus’ orbital

period?(b) It takes Mercury only 88 days to complete one orbit around the Sun.

Calculate Mercury’s mean distance from the Sun.

Satellite Orbits• A satellite is said to be in geosynchronous orbit if

it has a period of 1 day– Such satellites are used in communications and for

weather forecasting• For a satellite in a circular orbit:

€

v =GM

r

Example 12-11(a) Calculate the orbital speed of a satellite in circular orbit at an altitude

of 1,000 km above the surface of Earth.(b) Calculate this satellite’s orbital period.

Gravitational Potential Energy at any r

• Note that U = 0 at infinity, this is the common convention for problems of an astronomical scale– Doesn’t have to be, because only differences in U matter

€

U = −Gm1m2

r

U = mgh when h << R

Energy Conservation• The mechanical energy E of an object of mass m,

moving at speed v, at a distance from a planet of mass M is given by

• For a satellite in Kepler orbit, the total energy can be shown to be:

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E =K +U = 12mv

2 −GmM

r

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E = −GmM

2a

Orbital Speed

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v 2 =GM2

r−

1

a

⎛

⎝ ⎜

⎞

⎠ ⎟

Example 12-12An asteroid, headed directly toward Earth, has a speed of 12 km/s relative

to Earth when the asteroid is 10 Earth radii from Earth’s center. Neglecting the effects of Earth’s atmosphere on the asteroid, find the asteroid’s speed when it reaches Earth’s surface.

Example 12-13Venus orbits the Sun with a semi-major axis of 1.082×1011 m and an

eccentricity of 0.068. Calculate Venus’ orbital speed at periheion and aphelion.

Orbital Maneuvers • The radius at which a satellite follows a circular

orbital path is directly related to its speed• Moving to lower orbits requires you to use your

decelerating rockets – Slows satellite causing it to orbit in an elliptical orbit, a

second rocket burn is required to turn it into a circular orbit

• Moving to higher orbits requires you to use your accelerating rockets

The Hohmann Transfer to a lower orbit

The Hohmann Transfer to a higher orbit

Escape Speed• Escape speed is the minimum speed required to

leave a planet and to never return– To do this, K = 0 when r = ∞ (where U = 0)

€

ve =2GM

R

Example 12-14Suppose that you fired a cannonball straight upward with an initial

speed equal to one-half the escape speed. How far from the center of the Earth does this rocket travel before momentarily coming to rest? (Ignore air resistance in the Earth’s atmosphere)

Example 12-15What multiple of the energy needed to escape from Earth gives the

energy to escape from (a) the Moon, and (b) Jupiter? (the moon’s radius is 0.273 Earths and the moon’s mass is 0.0123 Earths,

while Jupiter’s radius is 11.21 Earths and its mass is 317.8 Earths)

Kepler Orbits

Orbital Shape Eccentricity (e) Semi-Major Axis (a)

Total Energy (E)

Circle e = 0 a > 0 E < 0

Ellipse 0 < e < 1 a > 0 E < 0

Parabola e = 1 none E = 0

Hyperbola e > 1 a < 0 E > 0

Black Holes & the Schwarzchild Radius

• Looking at the escape speed equation:you can seed that escape speed increases with increasing mass and decreasing radius

• If an object were to be compressed to a size small enough that the escape speed necessary were the speed of light, c then we have a black hole

€

ve =2GM

R

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RS =2GM

c 2

Example 12-16Mr. Bailey has a mass of about 110 kg. How small would you

have to compress him to turn him into a black hole?

Evidence for Black Holes

• Gravitational Lensing• Motion of stars near it