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Gravity & Orbits

  • Any two masses exert an attractive force on each other, known as the gravitational force.
  • Gravity is a non-contact force, known as a field force. Its effects are observed without the two objects coming into contact with each other.
  • A gravitational field describes the gravitational force a mass feels when placed at a particular point in space.
  • Inside a hollow sphere, the gravitational field is zero. Outside the sphere, you can treat the sphere as if its entire mass was concentrated at the center.
  • Inside a solid sphere, treat the sphere as if the mass inside the radius is all at the center. Outside the sphere, treat the sphere as if all the mass is concentrated at the center.
  • The velocity of a satellite is independent of the satellite’s mass.
  • Kepler’s Laws of Planetary Motion:
  • The orbits of planetary bodies are ellipses with the sun at one of the two foci of the ellipse.
  • If you were to draw a line from the sun to the orbiting body, the body would sweep out equal areas along the ellipse in equal amounts of time.
  • The ratio of the squares of the periods of two planets is equal to the ratio of the cubes of their semi-major axes. The ratio of the squares of the periods to the cubes of their semi-major axes is referred to as Kepler’s Constant.

Gravity & Orbits

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Objectives 0:07
  • Newton's Law of Universal Gravitation 0:45
    • Newton's Law of Universal Gravitation
  • Example I: Gravitational Force Between Earth and Sun 2:24
  • Example II: Two Satellites 3:39
  • Gravitational Field Strength 4:23
    • Gravitational Field Strength
  • Example III: Weight on Another Planet 6:22
  • Example IV: Gravitational Field of a Hollow Shell 7:31
  • Example V: Gravitational Field Inside a Solid Sphere 8:33
  • Velocity in Circular Orbit 12:05
    • Velocity in Circular Orbit
  • Period and Frequency for Circular Orbits 13:56
    • Period and Frequency for Circular Orbits
  • Mechanical Energy for Circular Orbits 16:11
    • Mechanical Energy for Circular Orbits
  • Escape Velocity 17:48
    • Escape Velocity
  • Kepler's 1st Law of Planetary Motion 19:41
    • Keller's 1st Law of Planetary Motion
  • Kepler's 2nd Law of Planetary Motion 20:05
    • Keller's 2nd Law of Planetary Motion
  • Kepler's 3rd Law of Planetary Motion 20:57
    • Ratio of the Squares of the Periods of Two Planets
    • Ratio of the Squares of the Periods to the Cubes of Their Semi-major Axes
  • Total Mechanical Energy for an Elliptical Orbit 21:57
    • Total Mechanical Energy for an Elliptical Orbit
  • Velocity and Radius for an Elliptical Orbit 22:35
    • Velocity and Radius for an Elliptical Orbit
  • Example VI: Rocket Launched Vertically 24:26
  • Example VII: AP-C 2007 FR2 28:16
    • Example VII: Part A
    • Example VII: Part B
    • Example VII: Part C
    • Example VII: Part D
    • Example VII: Part E

Transcription: Gravity & Orbits

Hello, everyone, and welcome back to www.educator.com.0000

I'm Dan Fullerton and in this lesson we are going to talk about gravity and orbits.0003

Our objectives include determining the force that one, spherical asymmetric mass inserted on another.0008

Determining the strength of the gravitational field at a specified point outside a spherically symmetric mass.0013

Describing the gravitational force inside and outside the uniform sphere.0020

Calculating how to feel that the surfaced depends on the radius and the density of the sphere.0024

Stating Kepler’s laws of planetary motion and using them to describe the motion of an object in orbit.0030

Finally, applying, conservation of the angular momentum to determine the velocity and radial distance at any point in that object’s orbit.0037

Let us start by taking a look at Newton’s law of universal gravitation.0044

The force of gravity, the force of attraction between any 2 objects with mass is given by this formula -G a constant, 0050

a fudge factor to make the units work out, × the mass of your first object × the mass of your second object ÷ the square of the distance between their centers.0059

R hat this means it is in the direction of r hat.0070

If we were to draw this, if we have object 1 and object 2 here, and we define from 1 to 2 as the R vector that means our unit vector in that direction, r hat is that direction.0072

The force of 1 on 2 was going to be opposite the direction of r hat, so it is going to attract 2 force of 1 on 2.0087

That is where the negative sign comes in.0098

Usually, when we are dealing with this, we will worry about magnitudes and use common sense to realize they are attracting each other 0100

but the negative sign has to do with this defining the direction from 1 to 2 as in the r hat direction.0105

Now that fudge factor G, the universal gravitational constant that is equal to 6.67 × 10⁻¹¹ N m² / kg².0112

If they want to make a plot of the gravitational force between 2 objects as a function of the distance between their centers, 0123

we would have this in versus square law, another place where as the distance increases between objects, 0130

we see the force go down as the square of the distance.0137

Let us do a quick sample here, find the magnitude of the gravitational force exerted on the earth by the sun given the mass of the earth, 0143

the mass of the sun, and the distance between them.0153

Let us start the force of gravity, it is going to be - GM 1M2 /r² in the direction of r hat, but since I'm worried about the magnitude, 0157

the magnitude of the gravitational force is just going to be 6.67 × 10⁻¹¹ N m² / kg².0169

G, the universal gravitational constant × our first math, the mass of the earth about 6 × 10²⁴ kg × 0183

the mass of the sun about 2 × 10³⁸ kg ÷ the square of the distance between them 1.5 × 10¹¹ N m².0192

Put all that in my calculator and I come up with the force of about 3.5 × 10²² N.0206

Let us do another example, the diagram below represents 2 satellites of equal mass A and B and circular orbits around the planet.0218

Compared to the magnitude of the gravitational force of attraction between A in the planet, 0227

the magnitude of the gravitational force of attraction between B and the planet is what?0232

Notice that B is twice as far from the center of the planet as A is.0237

Well, because it is twice as far and we have this inverse square law relationship, where the gravitational force is proportional to 1 / 0242

the square of the distance between the objects.0249

If we double the distance, we get ¼ the force, answer is C because of the inverse square law relationship.0252

As we are talking about gravity, let us take a look at the gravitational field.0265

Anywhere around the earth or any other object with mass are going to have some force of gravity.0270

To help us think about that, to visualize that as we talk about the gravitational field or the force that an object would feel if placed at that position.0276

What we are going to define is if the force of gravity is GM 1M2 /r² and if we are uniform gravitational field, we can write that as MG.0286

We can do a little simplification here and cancel out one of our masses to find that the gravitational field strength is GM 1 /r².0300

That would be the force an object would feel per unit mass.0312

As you look at these gravitational field lines, as they our more sparse you have less force.0316

As they are more dense, you have a greater gravitational field strength you would have more force per unit mass.0321

Now little G, we have been talking about is the acceleration due to gravity.0328

On the surface of the earth that is 9.8 m / s².0333

If we measure it this way and use these units we are going to come up with N / kg, m/ s² unit is equivalent to 1 N / kg.0337

Gravitational field strength g is the same thing as the acceleration due to gravity.0348

Let us do a quick example here, if we say that the radius of the earth is about 6,378 km, you can actually solve given the mass of the earth 0356

and the radius and find g is 9.8 m / s² or 9.8 N / kg.0367

Give it a try to confirm that reading.0377

Let us do another example, if you weigh 600 N on earth, what will you weigh on a planet with twice the mass of earth and half the radius of earth?0382

Of this is a Newton’s law of universal gravitation problem.0392

Then what happens if we double the mass?0398

If we double the masses, we are going to get twice the force, so we would double the force and you have 2 × what you had originally 0402

and if you half the radius because of the inverse square law you are going to quadruple the force.0413

At the radius and 2 and the denominator and it is squared, you will get 4 × what you have.0425

Put those two together, you are going to have an 8 × 8x increase.0432

8 × 600 N is going to give us a 4,800 N as your weight on this new planet.0437

Inside a hollow sphere the gravitational field is 0 and it is important to remember, inside a hollow sphere the gravitational field is 0.0453

Outside the hollow sphere you can treat it as if all of the mass were combined into one massive point at the very center, 0462

pretend its entire mass is concentrated at the center and then calculate the gravitational field. If we were to do that, we can go and graph it something like this. 0471

If this is our gravitational field strength and there is our R, inside the radius of the object we have 0 0484

and outside we have a gravitational field strength which is equal -GM 1 /R² in the direction of r hat.0496

What if we look at a solid sphere?0511

Outside a solid sphere, we said you can treat the spheres of all the masses at the center of the sphere.0514

Inside the sphere, however, you treat the sphere as if the mass inside the radius is all at the center, the only part that counts is what is inside the radius.0519

Outside the radius of where you are in that sphere does not count.0528

Only the mass inside the radius of interest counts.0532

As an example, let us say we want to know the gravitational field strength at some radius inside this solid sphere.0536

Let us draw another sphere inside and pretend that we are standing right there, somehow inside the solid sphere.0547

To start analyzing this problem, I’m first going to define a volume mass density ρ, 0554

which is the total mass ÷ the total volume or M / the volume of our entire sphere 4/3 π R³.0560

Defining the distance from the center to there is R.0573

Inside our sphere of interest here, what the volume enclosed in that sphere of interest is 4/3 π r³.0579

Our r is our radius out to the edge of our sphere of interest so then the mass enclosed by our sphere of interest 0589

is going to be our volume mass density × the volume of our sphere of interest which is going to be, we have for ρ, we have capital M /4/3 π R³.0598

For our volume enclosed we have 4/3 π r³.0617

Our mass enclosed is going to be, we can factor out 4/3, π, we are going to just get M, r³ /R³.0623

Looking for the gravitational field strength, G is - G × our mass enclosed/r², where our r is the distance to the edge to the radius of our sphere of interest 0640

is going to be - G /r² × our mass enclosed M r³/R³, which is just going to be - GM /R³ × r, where GM and R³ are constants.0655

What we have is a linear function of our radius and this is good for r is less than or equal to R.0677

If I wanted to make a graph of our gravitational field strength vs. our radius, there is our G and we are going to start at some 0 right at that center point.0688

We are going to have a linear increase and then when we get right to R, the edge of our total sphere are going to follow our inverse square log N.0700

Here in this region we are proportional to -r and in this region we are proportional to 1/r².0714

Let us talk about circular orbits.0725

Here we have an object M2 in a circular orbit about object M1.0728

The radius of our orbit is R, the force of gravity that is causing the centripetal force allowing it to move in the circle fg, 0734

at any point it has some tangential velocity V along with a rotational or angular velocity ω.0742

As we look at this to analyze that the net centripetal force which is always MAC, M2AC for our object 2 going around object 1, 0749

that has to equal the force of gravity because the force of gravity is what is causing that.0761

We know that the force of gravity by Newton's law of universal gravitation is GM 1M2 /r² and our centripetal acceleration we know is V² /r.0768

We could rewrite this as M2 V² /R must be equal to GM 1 M2 / R², which implies then that V² 0784

must be equal 2 GM1 /R or if we want just the velocity, the speed at any point in time, the velocity is the √ GM 1 /r.0806

It is in a circular orbit, the radius and the mass of the planet or whether we are orbiting about determines the speed.0820

Notice there is no dependence on the mass of the object that is orbiting.0829

Taking this a bit further, we can look at the period and frequency for a circular orbit.0836

Once around the entire orbit is the circumference or 2 π r and that has to equal the distance traveled which is going to be the velocity × the period, the speed × the period,0841

which implies then that period is 2 π r / V.0856

Due to our amazing physics skills, we just found out that the velocity for something in circular orbit is the √GM 1 /r.0863

Therefore, we can write that the period is 2 π r / our velocity or speed √GM 1 /r.0873

A little bit of rearrangement we can write that as 2 π √r / √GM 1 and all those square roots are making things a little messy.0887

If we square both sides, we can write this as T² = 4 π² r² and √r² is going to be an r.0899

In the numerator we will have r³ ÷ GM 1, that is known as Kepler’s third law for circular orbits.0909

What is pretty cool about this, is if we take the ratio of T² / r³, that is equal to 4π² / GM 1, that is known as Kepler’s constant.0931

For all the planets in our solar system, we have got the same M1 so we have roughly, assuming the roughly circular orbits and they are pretty close.0944

We have the same ratio of the square of their periods to the radius so that is Kepler’s constant.0952

It has a value of about 3 × 10⁻¹⁹ s² / m³.0959

We did a pseudo derivation of Kepler’s third law.0966

We could also look at the energy for objects moving in circular orbits.0970

Our total energy is the kinetic + the potential energy and we know for an object that is moving in a circular orbit we will first offer kinetic energy is ½ M2 V².0975

The gravitational potential energy is - GM 1M2 /r, this implies then that we could write the energy as equal to ½ M2 V² - GM 1M2 /r.0991

We also know that V² = GM 1 /r so we can write this as E = ½ M2 GM1 /r - GM 1 M2 /r, which is just going to give us - GM 1M2 /2r.1014

We have an expression for the total energy for an object in the circular orbit.1046

As we are doing our analysis of things in circular orbits, we can also talk about escape velocity.1068

Now escape velocity is a concept that describes the velocity an object needs while it is in caught by a gravitational field, 1073

the amount of speed it needs in order to completely escape the influence of that objects gravity.1082

We know practically, you cannot do that because you cannot get infinitely away from an object.1089

We can calculate how much energy that would require if you did everything perfectly.1095

It is a conservation of energy approach we are going to take here.1099

Let us start by looking at the total energy when we hit this point of being completely out of the influence of an object's gravity.1104

We will do that at the point where its gravitational potential energy is 0 and right at the point where it is kinetic energy is 0, where this completely stopped.1114

Anything past that, you would not need in order to escape that objects gravity that will be left over energy.1123

If we set that as our beginning condition and that means that ½ M2 V² - GM 1M2 /r must equal 0 or ½ M2 V² must equal GM 1M2 /r 1129

or solving for V² that is going to be 2 GM1 /r, which implies then that the escape velocity is going to be √2GM1 /r.1155

Since we have been talking about all these orbits, let us get into Kepler’s laws.1176

Kepler’s first law of planetary motion states that the orbits of planetary bodies or ellipses with a sun at 1 of the foe side of the ellipse.1182

We have 2 foe side, the sun is one of them, there is our first planetary law by Kepler.1191

The second one gets a little bit more mathematically in depth.1200

It says if you were to draw a line from the sun to the orbiting body, the body would sweep out equal areas along the ellipse in equal amounts of time.1206

That means, let us assume that we have an object in orbit, right now it is at 0.1 and at some point it gets here to 0.2.1214

As it does that, it sweeps out this amount of area, we will call that area 1.1221

At a later point in time, it is at P3.1228

If we let it go at the same amount of time that elapsed between P1 and P2, it will move from P3 to P4 and it will sweep out that this area of the ellipse.1231

As long as the time between P1 and P2 is the same as the time between P3 and P4, the area swept out in each of these cases is the exact same.1242

That is Kepler’s second law of planetary motion.1253

If we take a look at his third law and we also did this for circular orbits, it is a bit more complex.1257

It says the ratio of the squares of the periods, squares of the periods of 2 planets is equal to the ratio of the cubes of their semi major axis.1263

Instead of r³, we are doing this for a circular path, now we are going to use the semi major axis to find here for the ellipse.1272

This would be for an elliptical path T² /a³ = 4 π² / GM 1 or you might see this written as T² = 4π² a³ /GM 1.1283

The ratio of the squares of the period for the cubes of their semi major axis that is referred to as Kepler’s constant again.1302

There is Kepler’s third law of planetary motion, this time for elliptical orbits.1309

Let us take a look at total mechanical energy for elliptical orbit.1316

We are not going to go through the entire derivation, we did it for a circular orbit.1319

This would get a little more complex but the total energy is the kinetic + the potential which would is ½ M2 V² - GM 1M2 /r, when you are talking about circular orbits.1323

We get to elliptical orbits, when we do all the math and simplify it, we come out with the total energy as - GM 1M2 /2a, that is semi major axis instead of the radius of the circle again.1338

How about the velocity and radius from elliptical orbit?1355

As we look at this one, we have an object M2 and elliptical orbit about M1 and we are going to look into the couple points here.1359

As we analyze it, we can take a look starting with the derivative of the angular momentum about point P with respect to time has to equal the net torque about point B.1367

Assuming we have no external forces causing any net external torques, that has to be 0.1380

Therefore, our angular momentum about point P, I should say equal must be concert.1386

That means that the magnitude of the angular momentum about point P which will be at different points, if we look at that point A for example, 1401

that will be M2 velocity of A, radius at A, which would be the semi major axis × the sin θ A has to equal the same thing when it is at point B, M2 VB RB sin θ B.1410

And because the mass is the same, the masses cancel out, VA RA sin θ = VB RB sin θ B and that is true for any of the points on the ellipse.1429

Specifically, when you are at points A and B, the perigee point and what is known as the apogee point, at those points θ A = θ B those are 90°, 1440

which implies then that VA RA must equal VB RB, my simplification at those specific points.1453

Let us finish up with a couple problems here.1466

A rocket is launched vertically from the surface of the earth with an initial velocity of 10 km /s, what maximum height does it reached neglecting air resistance?1469

Some things that might be helpful, the mass of the earth, the radius of the earth, and we may not assume that the acceleration due to gravity is constant, 1479

it is going to be moving so far away from the surface of the earth.1486

The g is going to be changing in this problem.1490

We can solve this with the conservation of energy approach.1493

The initial kinetic energy + the initial gravitational potential energy must equal the final kinetic + final gravitational potential energies.1497

This implies then that ½ × M2, the mass of our object × the square of its velocity - GM 1M2 / the radius of the earth.1508

Its gravitational potential energy when it is on the surface of the earth because now our sending arbitrary 0 point is infinity has to equal to, 1522

its highest point kinetic is 0 and its gravitational potential energy will be - GM 1M2 /r.1531

So then this implies, if we solve for the square of the velocity V² - 2 GM 1 /E = -2 GM1 /R.1542

Notice that M2 has completely have been eliminated from the equation but we kind of expected it.1558

A little bit more rearrangement, 1 /r = -V² / 2 GM1 + 2 GM 1/2 GM 1 RE, radius of the earth which implies that 1565

1 /r = - V² / 2 G M1 + 1 / the radius of the earth, 1584

which implies that 1 /r = we have our negative square the initial velocity 10,000 m / s² ÷ 2 × G 6.67 × 10⁻¹¹ N m² / kg² × the mass of the earth 1594

6 × 10²⁴ kg + 1 /the radius of the earth 6.37 × 10⁶ m, which implies then that r is approximately equal to 3.12 × 10 ⁺7m.1615

Now it is important here to realize that, that is the total distance from the center of the earth.1635

If we want to know the maximum height it reaches, how far above the surface of the earth, what we have calculated here, 1640

if there is earth, there is radius of the earth, we have calculated how high it goes there.1648

We are really want to know this distance so that means we have to go a little bit further, the height above the surface of the earth, 1655

let us call that H = the radius - the radius of the earth which is going to be our 3.12 × 10⁷ m - the radius of the earth 1664

6.37 × 10⁶ m or 2.48 × 10 ⁺7m above the surface of the earth.1679

Let us finish out by doing an old AP free response problem.1692

We are going to start with a 2007 exam free response question number 2.1697

We are looking at the Mars global surveyor, entered its final orbit about Mars and gives us some data about that, 1705

we are asked to calculate the radius of the global surveyor's orbit.1711

For part A, it is going in some orbit where we have a radius of Mars of 3.43 × 10⁶ m /s, I should probably should put that in red for Mars.1716

The period it tells us is 118 min or 7080 s, our mass is 930 kg and our velocity is 3400 m/s, calculate the radius then.1734

Period is the distance traveled ÷ the velocity so that is going to be 2 π × the radius ÷ the speed of which implies then 1754

that r is going to be equal the VT / 2 π which is 3400 m /s × our period 7080 s / 2 π or I come up with a radius of our orbit of about 3.83 × 10 ⁺6m.1764

Looking at part B, calculate the mass of Mars.1792

We can go back to our Newton’s second law to write that the net centripetal force which is going to be MV² /r must equal the gravitational force, 1797

Newton’s law of universal gravitation because the gravitational force is providing that centripetal force GM 1M2 /r², 1811

which implies then that V² is going to be equal 2 G × the mass of Mars ÷ R,1821

which implies then that the mass of Mars must equal the radius × to square root of velocity ÷ 1831

the gravitational constant which is 3.83 × 10 ⁺6m, we just found that out up there.1838

The square of our velocity is going to be 3400 m /s² ÷ the universal gravitational constant 6.67 × 10⁻¹¹ N m² / kg².1848

It comes out to be about 6.64 × 10²³ kg.1863

For C, it is asking us to find the total mechanical energy of the global surveyor in that orbit.1876

Our total energy is the kinetic energy + the potential energy which is going to be ½ MV² – G mass of the satellite, 1885

Mass of Mars over the radius which will be ½ × our mass 930 kg × our speed 3400² – G 6.67 × 10⁻¹¹ × 1897

the mass of our satellite 930 × the mass of Mars 6.64 × 10²³ ÷ 3.83 × 10⁶.1915

I come up with a total energy then, when I put that in my calculator of about -5.38 × 10⁹ joules.1930

Let us take a look at D, if the global surveyor was replaced in the lower circular orbit, will the new orbital period be greater than or less than a given period?1947

That is definitely going to be less than.1957

The velocity must go up, the shorter circumference to travel so the distance it is traveling is going down.1961

And if velocity is displacement /time, distance traveled / time, and time is displacement/ velocity.1968

We have a distance going down and the velocity going up, both those factors lead to a shorter time period /orbit.1977

I might explain that out in a little more detail and sentences but that is the idea,1992

And for part E, we have a question about the global surveyor entered an elliptical orbit, if the speed of its closest approach was 3400 m/s, 1997

find the speed at the furthest point of its orbit.2012

We can do that by conservation of spin angular momentum and state that the mass of our satellite × the velocity position 1 × the radius of position 1 = 2015

the mass of the satellite × the velocity of position 2 × the radius of position 2 or V1 r1 = V2 r2.2026

If we want the velocity at that new point, V2 is just V1 r1/r2 which would be our 3400 m/s × our radius 3.71 + 34.3 × 10⁵ m.2037

I'm just going to use a short hand, we do not need to put all of the thousands in there since they are all going to factor out here, over 4.36 + 34.3 × 10⁵ m.2057

We will have this ratio × 3400, gives us a velocity of about 3343 m /s.2077

Hopefully, that gets you a great start on gravity and orbits.2090

Thank you for joining us today at www.educator.com.2093

I look forward to seeing you again real soon and make it a great day everybody. 2095