Ch 13 - Universal Gravitation

Newton’s Law of Universal Gravitation

In Ch 1, you are introduced to the formula

F = G Mm/r² for use in unit analysis.

Now that you know force is also measured in Newtons, what are the units for “G”?

Using the force that the Earth exerts on a apple and on the moon, Newton provided strong evidence of the inverse square relationship of F = “G” M m / r²

As you see below…the required centripetal acceleration required to keep the moon in orbit about the earth corresponds with the Force due to gravity caused by the Earth-Moon system

F = “G” M m / r²

N = “G” kg kg /m²

“G” = N m²/ kg²

à G = 6.673 x 10^-11 Nm²/kg²

Let’s apply Ch 6, Newton’s 3^rd law. For every action there is an equal and opposite reaction.

If the Earth exerts a force of, F, on the moon, how much force does the moon exert on the Earth?

+F_EarthOnMoon = -F_MoonOnEarth

mappleg	^=	“G” ME mapple / rE^2
mmoona		“G” ME mmoon /rEtoM^2

g	^=	1 / rE^2
a		1 /rEtoM^2

9.8	^=	1/(6.37x10^6)^2
a		1 /(3.84 x 10^8)^2

a_c = 2.75 x 10^-3 m/s²

where ma_c is a centripetal force since we know the moon orbits the Earth and doesn’t fly off in a straight line

We also know the period of the moon

T_moon = 27.4day(24hr/day)*(3600s/day)

a_c = v² / r

a_c = (2πr/T)² / r

a_c = 4π²r/T²

a_c = 4π²3.84 x 10⁸ /(2.36x10⁶)²

a_c = 2.72 x 10^-3 m/s²

Example (13.1 and 13.3)

Let’s calculate the mass of the Earth.

m g = G Mm / r²

9.79 = 6.673x10^-11 M / (6.4x10⁶)²

M = 6 x 10²⁴ kg

Measuring the Gravitational Constant

By varying the distance of the large black bar bell and varying the masses and measuring the induced force using the twist of the wire, “G” was able to be calculated in 1798.

Free-Fall Acceleration and the Gravitational Force

Kepler’s Laws and the Motions of Planets

1^st Law: All planets move in elliptical orbits with the sun at one focus

How to make an ellipse?

Ø Choose any two points, called foci

Ø Attach both ends of a long string at the foci

Ø Take a pencil and draw on the inside while keeping the string taut

Any corresponding r₁ + r₂ are constant.

Notes:

The elliptical shape is a direct result of the inverse square law

Measurements of ellipse

Eccentricity à e

e = c / a

important notes:

As c à 0; e = 0

When e = 0 the ellipse is a circle

a à semi major axis

2a à major axis

b à semi minor axis

2b à minor axis

c à used to measure eccentricity

2^nd Law: The radii vector sweeps out equal areas in equal times

See below for better sketch

average velocity

v_ave = Δx/Δt

v_ave = Δr/Δt

Δr = v_ave Δt

instantaneous v

v = dx / dt

v = dr / dt

dr = v dt

As we can see in the sketch to the right, the force vector is parallel to the radii vector.

Recall from Ch 11 à t = r x F

Since r || F, then t = 0 Nm

(rules of cross products)

Area of triangle is

½ base x height

if angle between r₁ & r₂ is infinitely small, then Δr à dr

and

Area of triangle is

dA = ½ r x dr

dA = ½ r x v dt

dA = ½ r x v dt (m/m)
dA = ½(r x mv) dt / m

dA = ½ ( L ) dt / m

dA = L dt / 2m

Recall from Ch 11 à t = dL / dt

Another way to look at this…

the more time you apply a torque, the greater the angular momentum

dt t = dL

Since t = 0 Nm we know that angular momentum is conserved

Recall from Ch 11 à L = r x p

L = r x p

L = r x mv

dA / dt = L / 2m since L is conserved; m is a constant

dA / dt = L / 2m = constant

Since dA / dt = constant, we get Equal Areas in Equal Times à this is the proof for Kepler’s 2^nd law

3^rd Law: The square of the Period, T, is proportional to the semi-major axis of the elliptical orbit or simply

3^rd Law: T² = k a³ (k is constant and changes for each body)

F = GMm/r² F_c = mv²/r

Since gravitational attracting is supplying the centripetal force

a = r, for circular orbits

for elliptical

T²/a³ = k_sun

Kupiter belt objects: Pluto

30 to 50 Au

GMm/r² = mv²/r

GM/r = v² where (v = 2πr/T)

GM/r = (2πr/T)²

GM/r = 4π²r²/T²

T²/r³ = 4π²/GM

T²/r³ = k_object

k_sun = 4π²/GM_sun

k_earth = 4π²/GM_earth

use mass of sun and

k_sun = 2.97 x 10-19 s²/m³

Example

As thermonuclear fusion proceeds in its core, the Sun loses mass at a rate of 3.64 ´ 10⁹ kg/s. During the 5000 year period of recorded history, by how much has the length of the year changed due to the loss of mass from the Sun?

F = m(v²/r)

v = 2πr/T

F = m((2πr/T)²/r)

F = 4m π²r/T²

F = GMm/r²

GMT² = 4 π² r³

GMT^1/2 = 4π²(L/2mπ)^3/2

The rate of change is (derivative)

GM(½T^-1/2dT/dt) + GT^1/2 dM/dt = 0

dT/dt = -dM/dt (2 T / M)

∆T/∆t ≈ -dM/dt (2 T / M)

∆T ≈ - ∆t (dM/dt) (2 T / M)

∆T ≈-5000(3.64´10⁹)(2(1)/1.991x10³⁰)

∆T ≈ 1.83 x 10^-17 yr²/sec

3.16x10⁷ sec in one year

∆T ≈ 0.0183 sec

L = r x p

L = r x mv

r ≈ ┴ p

L = mvr

L = m(2πr/T)r

L = 2m π r²/T

r² = LT/2mπ

The Gravitational Field

F = mg

F = GMm/r²

from Earth’s point of view

Every object creates a gravitational field surrounding it, which all other objects are attracted.

mg = GMm/r²

Since gravitational attracting is supplying the gravitational force

Gravitational Potential Energy

If you raise an object, you do work on the object, but gravity also does work on the object in the opposite direction.

Work_gravity = - F Δr

Work_gravity = -GMm/r² dr

Work_gravity = -GMm ò 1/r² dr

Work_gravity = D GMm/r

Work_gravity = GMm(1/r_final –1/r_init)

r_final = ¥

Work_gravity = -GMm(1/r)

You raise an 1 kg object one meter in the air, you do 10 Nm of work raising the object

Work = mg * Δy

Work = 10 N * 1 m

Work_gravity = negative of you

Work_gravity = -10 Nm

But what if the object is

raised 10⁶ meters.

Gravity varies in this range.

Thus the force varies

(not constant)

30 AU’s past the Kupiter belt we are on a iron mining expedition and find an almost formed asteroid. It’s still in 3 roughly spherical pieces, each of 10⁸ kg. About how much energy must be expended to move them to 1000 meter equilateral triangle positions?

ρ_iron-rock = 5000 kg/m³

U_total = U₁₂ + U₂₃ + U₁₃

U_total = -G(m₁m₂/r₁₂ + m₂m₃/r₂₃ + m₁m₃/r₁₃)

U_total = -3G(m²/r) (m₁ = m₂ = m₃; r₁ = r₂ = r₃)

V = m/ρ = 10⁸/5000

20000 = 4/3πr³

r = 36 m

U_total = -3G(m²/r) ¥ to 36 m

U_@36m = -3(6.7x10^-11)(10^8*2/36)

U_@36m = -55,800 Joules

U_total = -3G(m²/r) from ¥ to 1000

U_to1000 = -3(6.7x10^-11)(10^8*2/10³)

U_to1000 = -2010 Joules

So it takes about 53790 Joules of energy to pull them apart

13.7 Energy Considerations in Planetary and Satellite Motion

The total kinetic and gravitational potential energy of an object is ½mv² + Work_gravity to raise it to it’s current position

For an object in orbit, what supplies the inward (centripetal) force?

Ans: Gravity

E is total energy

E = U + K

E = -GMm/r+½GMm/r

E = -½ GMm/r

in circular orbits

E = -½ GMm/a

in elliptical orbits

To calculate escape velocity, v_esc, our initial kinetic energy must be enough to transfer to raise the object to desired height…once achieved the KE at the top of the path will be zero.

U_bottom + K_bottom = U_top + K_top

-GMm/r_Earth + ½mv_esc² = -GMm/r_max+ 0

v_esc² = 2GM/r_Earth - 2GM/r_max

v_esc² = 2GM(1/r_Earth - 1/r_max)

We know that r_max is essentially ¥, so the 1/r_max = 0 and the equation simplifies to

mv²/r = GMm/r²

mv²/r(½r) = GMm/r² (½r)

½ mv² = ½ GMm/r

K = ½ GMm/r

Note K = ½ of Potential energy

U = -GMm/r (from 13.6, above)

Extra pictures