Lecture Notes on Gravitational Forces and Orbits in PHYS-2010: General Physics I - Prof. D, Study notes of Physics

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PHYS-2010: General Physics I

Course Lecture Notes

Section IX

Dr. Donald G. Luttermoser East Tennessee State University

Edition 2.

Abstract

These class notes are designed for use of the instructor and students of the course PHYS-2010: General Physics I taught by Dr. Donald Luttermoser at East Tennessee State University. These notes make reference to the College Physics, 8th Edition (2009) textbook by Serway and Vuille.

IX–2 PHYS-2010: General Physics I

center m

M (^) p

r

R (^) p h

Note that r = Rp + h (radius of sphere, i.e., planet, plus the height above the sphere). Example IX–1. Problem 7.40 (Page 223) from the Ser- way & Vuille textbook: Two objects attract each other with a gravitational force of magnitude 1. 00 × 10 −^8 N when separated by 20.0 cm. If the total mass of the two objects is 5.00 kg, what is the mass of each? Solution: Let the individual masses be represented by m 1 and m 2 and the total mass of the two objects be represented as M = m 1 + m 2 (= 5.00 kg). Then we can express the second mass as m 2 = M − m 1. The distance between the two masses is r = 20.0 cm = 0.200 m, the gravitational force between them is F = 1. 00 × 10 −^8 N, and the gravitational constant is G = 6. 67 × 10 −^11 N m^2 /kg^2. Now make use of Newton’s Law of Gravitation: F = Gm 1 m 2 r^2 =^

Gm 1 (M − m 1 ) r^2 F r^2 = Gm 1 (M − m 1 ) = GMm 1 − Gm^21 0 = Gm^21 − GMm 1 + F r^2 Algebra has a well know solution to the quadratic equation of the form ax^2 + bx + c = 0 , as x = −b ±

b^2 − 4 ac 2 a

Donald G. Luttermoser, ETSU IX–

Here, x = m 1 , a = G, b = −GM, and c = F r^2 , so

m 1 = GM^ ±

G^2 M^2 − 4 GF r^2 2 G =

GM ± G

√ M^2 − 4 F r^2 /G 2 G =

 M ±

√√ √√ M^2 − 4 F r^2 G

 

  5 .00 kg ±

√√ √√ √(5.00 kg)^2 − 4(1.^00 ×^10

− (^8) kg)(0.200 m) 2

1. 67 × 10 −^11 N m^2 /kg^2

 

[ 5 .00 kg ±

√ 25 .0 kg^2 − 24 .0 kg^2

]

[ 5 .00 kg ±

√ 1 .0 kg^2

] . As such, m 1 either equals (4.0 kg)/2 = 2.0 kg or (6.0 kg)/2 = 3.0 kg (either answer is correct). If we chose m 1 = 3.0 kg, then m 2 = M − m 1 = 5.00 kg − 3.0 kg = 2.0 kg. Hence the solution is m 1 = 3.0 kg and m 2 = 2.0 kg.

1. Why do objects fall independent of their mass on the Earth’s surface? a) Gravity: Fg = G M⊕ m R^2 ⊕

b) Motion: F = m a = m g.

c) Set Fg = F , then G M⊕ m R^2 ⊕ = m g or g = G M⊕ R^2 ⊕

. (IX-3)

Donald G. Luttermoser, ETSU IX–

=⇒ the speed of light’s accuracy is known to 9 significant digits!

c = 2.99792458(1) × 108 m/s.

d) Future space-based experiments (in free-fall and a vac- uum) should increase the accuracy in the measurement of G!

1. The farther we get from the Earth’s center, the smaller the ac- celeration due to gravity:

g =

G M⊕

r^2

, (IX-4)

where r ≡ distance from Earth’s center. a) When g is measured on the Earth’s surface (or some other planetary surface), it is called the surface gravity.

b) When g is measured elsewhere (i.e., not on a planetary surface), it is called the acceleration due to gravity, and if the object is in free-fall, it also is often called the free-fall acceleration.

Example IX–2. Mt. Everest is at a height of 29,003 ft (8840 m) above sea level. The greatest depth in the sea is 34,219 ft (10, m). Compare the Earth’s surface gravity at these two points. Solution: Let he be the height of Mount Everest and hs be the greatest depth of the sea. Using Eq. (IX-4) and the fact that the radius of the Earth at sea level at the Earth’s equator is R⊕ = 6. 378077 × 106 m, we get the distances for the highest and lowest points of the surface from the center of the Earth as rh = R⊕ + he = 6. 378077 × 106 m + 8. 840 × 103 m

IX–6 PHYS-2010: General Physics I

= 6. 386917 × 106 m rl = R⊕ − hs = 6. 378077 × 106 m − 1. 0430 × 104 m = 6. 367647 × 106 m

Which gives the surface gravities of

gh =

G M⊕

r^2 h

(6. 6730 × 10 −^11 N m^2 /kg^2 )(5. 9763 × 1024 kg) (6. 386917 × 106 m)^2 = 9 .7762 m/s^2 ,

gl =

G M⊕

r^2 l

(6. 6730 × 10 −^11 N m^2 /kg^2 )(5. 9763 × 1024 kg) (6. 367647 × 106 m)^2 = 9 .8355 m/s^2 ,

and gl − gh g

× 100% =

× 100% = 0.605% ,

the surface gravity of the lowest point on the Earth’s surface is a little more than half of a percent larger than at the highest point.

B. Kepler’s Laws of Planetary Motion.

1. Johannes Kepler (1571 – 1630) was a German mathematician and astronomer who used Tycho Brahe’s observations of Mars to derive the 3 laws of planetary motion. The data showed that the Copernican model of heliocentric (Sun-centered) solar system was correct, except that the planets move in elliptical and not circular paths around the Sun as Copernicus had assumed.
2. The laws: a) Law 1: The orbit of a planet about the Sun is an ellipse with the Sun at one focus. The so-called elliptical or- bit. The equation for the ellipse in Cartesian coordinates

IX–8 PHYS-2010: General Physics I

v) The distance from the center to either focus is given by

a^2 − b^2.

vi) The ellipse is just one type of conic section. If a = b, Eq. (IX-6) gives e = 0 and we have a circular orbit =⇒ a second type of conic section (i.e., a circle).

vii) If we let ‘a’ get bigger and bigger such that a  b, then Eq. (IX-6) gives e ≈

a^2 /a = a/a = 1 as a → ∞. When this happens, we have a parabolic orbit. Such an orbit is said to be open (both cir- cular and elliptical orbits are closed) and never return. A parabolic orbit is achieved when the ve- locity of a satellite just equals the escape velocity, vesc.

viii) There also are orbits that are “more open” than parabolic orbits =⇒ the so-called hyperbolic or- bits. These orbits have e > 1 and can be achieved if v > vesc.

b) Law 2: A line joining a planet and the Sun sweeps out equal areas in equal amounts of time (law of equal areas ).

i) This means that planets move faster when closer to the Sun in its orbit than when it is farther away.

ii) Objects in very elliptical orbits don’t stay near the Sun for a very long time =⇒ comets.

iii) Perihelion: Point on an orbit when a planet is closest to the Sun (rp = perihelion distance).

Donald G. Luttermoser, ETSU IX–

iv) Aphelion: Point on an orbit when a planet is farthest from the Sun (ra = aphelion distance).

v) The perihelion and aphelion of a solar orbit can be determined from the semimajor axis and the ec- centricity with

rp = a(1 − e) (IX-7) ra = a(1 + e) , (IX-8)

also note that rp + ra = 2a. (IX-9)

Sun A (^1)

t (^1) A (^2)

t 2

if A 1 = A 2 (area), then t 1 = t 2 (time)

rp ra

c) Law 3: The square of the orbital period (T ) of any planet is proportional to the cube of the semimajor axis (a) of a planet’s orbit about the Sun (harmonic law ):

T 2 ∝ a^3. (IX-10)

Donald G. Luttermoser, ETSU IX–

i) K = 4 π^2 G M = 2. 97 × 10 −^19 s^2 /m^3.

ii) K is independent of the planet’s mass =⇒ K only depends upon the larger, central object’s mass.

iii) T 2 ∝ a^3 is valid whether the orbit is circular (where a = r, the orbital radius) or elliptical (where ‘a’ is used intact), though the elliptical orbit solu- tion has a slightly different form for the K con- stant. The proof for elliptical orbits requires ad- vanced calculus (hence we will not show it here).

1. We can base other planets in solar system with respect to the Earth: T (^) p^2 T (^) ⊕^2

K a^3 p K a^3 ⊕

a^3 p a^3 ⊕

. (IX-12)

Since T⊕ = 1 yr and a⊕ = 1 A.U., Eq. (IX-12) becomes

T (^) p^2 1 yr^2

a^3 p 1 A.U.^3

or T (^) yr^2 = a^3 AU. (IX-13)

Example IX–3. The Voyager 1 spacecraft has just passed the 100 A.U. mark in its distance from the Sun. At this time, the Death Star from an evil galactic empire intercepts Voyager and figures out which planet sent it based on the gold record that was included on the spacecraft. The Death Star alters its course and starts to orbit the Sun from that point and sets its heading such that it will reach perihelion at the Earth’s location. Calculate the following about its orbit: (a) the semimajor axis (in A.U.); (b) the eccentricity, and (c)

IX–12 PHYS-2010: General Physics I

the length of time (in years) it will take to get to Earth. Solution (a): We are given that rp = 1.00 A.U. and since it starts its free fall orbit from the position of Voyager 1, that position marks it aphelion position, ra = 100 A.U. (note that we do not have to convert to SI units for this problem). Using Eq. (IX-9) we can easily calculate the semimajor axis of the Death Star’s orbit:

a = rp + ra 2

1.00 A.U. + 100 A.U.

= 50 .5 A.U.

Solution (b): Using Eq. (IX-7) and solving for e, we get,

e = 1 − rp a

1.00 A.U.

50.5 A.U.

nearly a parabolic orbit! Solution (c): First we need to calculate the period of a complete orbit from Kepler’s 3rd law, then the amount of time it will take to go from aphelion to perihelion is one-half that period. As such, ( (^) T T⊕

) 2

( (^) a a⊕

) 3

( 50 .5 A.U.

1 .00 A.U.

) 3 = (50.5)^3 = 1. 29 × 105 T 1 .00 yr =^

1. 29 × 105

T = 359 yr Since this is the full period of the Death Star’s orbit, the time to get to Earth will be

t =^1 2 T = 180 yr.

IX–14 PHYS-2010: General Physics I

ii) Now if we define the initial PE◦ as 0 at the center of the gravitating body (so r◦ = 0), Eq. (IX-16) becomes

PE = −

( (^) G M m r^2

) r

= −G M m r

iii) The calculus solution to the differential equation above (Eq. IX-15) would have given exactly the same answer.

iv) As such, the most general form of the potential energy of a mass in a gravitating field is

PE = − G M m r.^ (IX-17)

c) With this general form of the gravitational potential en- ergy, we can see how the potential energy equation near the Earth’s surface (Eq. IX-14) arises. i) Let’s say we have a projectile that we launch from the ground. While on the ground, r = R⊕ (the radius of the Earth), which gives a potential energy of PE = − G M⊕ m R⊕

. (IX-18)

ii) Now, when the projectile reaches its highest point above the ground, h, it is a distance of r = R⊕ + h from the center of the Earth. At this point, it has a potential energy of

PE = − G M⊕ m R⊕ + h

. (IX-19)

Donald G. Luttermoser, ETSU IX–

iii) The change in potential energy between these two points is ∆PE = −G M⊕^ m R⊕ + h − −G M⊕^ m R⊕ = G M⊕ m

R⊕

R⊕ + h

)

= G M⊕ m

  R⊕^ +^ h R⊕ (R⊕ + h) −^

R⊕

R⊕ (R⊕ + h)

 

= G M⊕ m

  R⊕^ +^ h^ −^ R⊕ R⊕ (R⊕ + h)

 

= G M⊕ m

  h R⊕ (R⊕ + h)

  (^).

iv) If R⊕ is much greater than h (which it will be for experiments near the Earth’s surface), h  R⊕. As such, R⊕+h ≈ R⊕ and the equation above becomes

∆PE ≈ G M⊕ m

  h R⊕ (R⊕)

  (^) = G M⊕ m

  h R^2 ⊕

 

G M⊕ m h R^2 ⊕

v) Now, remembering our defining equation for sur- face gravity (e.g., Eq. IX-3):

g =

G M⊕

R^2 ⊕

we use this in the potential equation we just wrote: ∆PE = m G M⊕ R^2 ⊕ h = m g h ,

hence we have proven Eq. (IX-14) from first prin- ciples.

1. We can now use this general potential energy equation to figure out high trajectory, orbital, and space trajectory problems. First,

Donald G. Luttermoser, ETSU IX–

v ◦^2 = 2 GM⊕

R⊕

r

)

v◦ =

√√ √√ 2 GM⊕

R⊕

r

)

. (IX-20)

d) Now if we once again define g to be the acceleration due to gravity at the Earth’s surface (i.e., surface gravity), we can rewrite Eq. (IX-3) to read

G M⊕ = g R^2 ⊕.

e) Plugging this into Eq. (IX-20) gives

v◦ =

√√ √√ 2 gR^2 ⊕

R⊕

r

) , (IX-21)

and replacing r by R⊕ + h, we get

v◦ =

√√ √√ 2 gR^2 ⊕

R⊕

R⊕ + h

)

√√ √√ √ 2 gR^2 ⊕

  R⊕^ +^ h R⊕ (R⊕ + h)

R⊕

R⊕ (R⊕ + h)

 

√√ √√ √ 2 gR^2 ⊕

  R⊕^ +^ h^ −^ R⊕ R⊕ (R⊕ + h)

 

√√ √√ √ 2 gR^2 ⊕

  h R⊕ (R⊕ + h)

 

√√ √√ 2 gR⊕

( (^) h R⊕ + h

)

v◦ =

√√ √√ 2 ghR⊕ R⊕ + h

. (IX-22)

f) Finally, one can immediately see that if h  R⊕, then R⊕ + h ≈ R⊕ and Eq. (IX-22) becomes

v◦ ≈

√√ √√ 2 ghR⊕ R⊕^ =^

√ 2 gh.

IX–18 PHYS-2010: General Physics I

As can be seen, this equation (as written above) is just an approximation to a more general equation (i.e., Eq. IX-22).

Example IX–4. Problem 7.45 (Page 223) from the Serway & Vuille textbook: A satellite of mass 200 kg is launched from a site on the Equator into an orbit at 200 km above the Earth’s surface. (a) If the orbit is circular, what is the orbital period of this satellite? (b) What is the satellite’s speed in orbit? (c) What is the minimum energy necessary to place this satellite in orbit, assuming no air friction?

Solution (a): The radius of the satellite’s orbit is

r = R⊕ + h = 6. 38 × 106 m + 200 × 103 m = 6. 58 × 106 m ,

m = 200 kg is the satellite’s mass, and M⊕ = 5. 98 × 1024 kg is the Earth’s mass. The orbital velocity will just be the tangen- tial velocity of the circular orbit. Since the gravitational force provides the centripetal acceleration, we have

Fc = Fg m

  v

(^2) orb r

  (^) = GM⊕m r^2 v^2 orb = GM⊕ r vorb =

√√ √√ GM⊕ r

=

√√ √√ √ (6.^67 ×^10 − (^11) N m^2 /kg^2 )(5. 98 × 1024 kg)

1. 58 × 106 m = 7. 79 × 103 m/s.