Physics 121 Concepts - SummaryLast updated March 11, 2010, 1:40 pm

Note: Although I have done my best to check for typos and list the formulas correctly, you should verify the formulas are correct before using them. - Dr. Nazareth

Know/Understand the information in Appendices of your textbooko Powers of Ten and Scientific Notationo Significant Figureso Algebrao Exponents and Logarithmso Geometry and Trigonometryo Sine, cosine, tangent only apply to right triangles (where one of the angles = 90°)o Tables of typical values, planetary data, elements, isotopes

Units (1.2) Dimensional Analysis (1.3) Conversion of Units (1.5)

o Units should always be written out and carried throughout calculations.o Units obey the same algebraic rules that numbers do because they are mathematical

entities.o In a conversion, if the units do not combine algebraically to desired result, then the

conversion was not done correctly. Order of Magnitude Calculations Scalars and Vectors (1.7)

o A scalar quantity is described completely by its size or magnitude.o A vector quantity has a magnitude (size or strength) and a direction.

Problem Solving in Physics (1.8)o Read the problem carefullyo Sketch the systemo Visualize the physical processo Strategizeo Identify appropriate equationso Solve the equations (use variables to solve and input numerical values at the end)o Check your answer to see if it makes sense/is reasonable/is what you expected

Position, Distance, and Displacement (2.1)o Start by defining a coordinate systemo Distance – total length of travelo Position – where you are relative to origino Displacement - a vector that points from an object’s initial position (usually taken

to be the origin or 0,0) to its final position. The magnitude (size) of the displacement vector equals the shortest distance

between the initial and final positions. Remember there are positive and negative directions (point opposite way)

Average Speed and Velocity (2.2)

o Average speed = distance traveled/elapsed time Speed is a scalar quantity = magnitude only ex: car travels 200 m in 10 sec; average speed = 20 m/s

o Average velocity = displacement/time vector quantity = magnitude and direction direction is the same as the displacement (can be positive or negative) Graphical interpretation of average velocity – x-versus t graph

o Average velocity is slope of straight line connecting two points corresponding to given time interval.

Instantaneous Velocity (2.3)o Instantaneous velocity – indicates the speed of the object and the direction of

motion at any moment in time instantaneous velocity can change in magnitude or direction or both vector quantity = magnitude and direction direction is the same as the displacement (can be positive or negative) In this class we use the word velocity to mean the instantaneous velocity. Graphical interpretation - – x-versus t graph

o Instantaneous velocity is slope of tangent line at given instant of time Acceleration (2.4)

o Acceleration is the change in velocity over a certain interval of time.o Acceleration changes the velocity (in magnitude, direction, or both).o Acceleration can speed you up or slow you down (often called deceleration).o Constant velocity means the acceleration is zero and visa versa.o Average acceleration = change in velocity/elapsed time

vector quantity = magnitude and direction direction is the same as the change in velocity (can be positive or negative)

o Instantaneous acceleration – indicates the acceleration of the object and the direction of motion at any moment in time instantaneous acceleration can change in magnitude or direction or both vector quantity = magnitude and direction direction is the same as the change in velocity (can be positive or negative) In this class we use the word acceleration to mean the instantaneous

acceleration.o In this class, we usually assume the acceleration is constant.o Make sure you know the difference between initial velocity, final velocity, and

average velocity. The final velocity only equals the average velocity when the acceleration is zero.

Equations of Motion for Constant Acceleration (2.5)v = v0 + at Assumptions x = displacementvave = ½(v0 + v) & a = acceleration (= constant)x = x0 + ½(v0 + v)t Definitions v = final velocity at time tx = x0 + v0t + ½ at2 v0 = initial velocity at time t0 = 0v2 = v0

2 + 2aΔx t = time elapsed since t0 = 0 Applications of the Equations of Motion (2.6) Strategy for Solving Physics Problems

1. Read the problem carefully.

2. Visualize the motion.3. Draw a picture to represent the situation.4. Choose a convenient coordinate system if one is not specified for you to use. Decide

positive and negative direction.5. Write down the values given to you and identify what they represent (e.g., the 2 m/s

given is the final velocity).6. Identify and write down what variables/quantities you are being asked to determine.7. Identify the physics of the problem. This will point you to the equation or set of

equations that are suitable for the problem.8. Strategize. Look at what variables/quantities you have been given and think about

which equations you know that might be applied to the problem. If you think you don’t have enough known variables, remember that the motion of two or more objects might be related by a common variable like time or the distance between the two objects.

9. Use algebra to solve the equation(s) for the unknown using variables. After the unknown variable is solved for, then stick the actual numbers (with units!) into the equation to solve for the final numerical answer. Remember to include your units in the final equation and make sure they come out right.

Freely Falling Objects (2.7)o Free-fall – idealized motion where air resistance is negligible and acceleration

approximately constanto Acceleration of a freely falling body called acceleration due to gravity

Magnitude at Earth’s surface = g = 9.80 m/s2 = 32.3 ft/s2

Always directed downward toward center of the Eartho In absence of air resistance, book and feather fall with same acceleration and land

at same timeo Any object moving under influence of gravity only is a freely falling object

motion can be downward or upward always experiences downward acceleration due to gravity

o time symmetry – object thrown upward takes same time to reach maximum height as to return from there to starting point

at maximum height, the velocity of the object thrown upward is 0 (zero)

Scalars and Vectors (3.1)o A scalar quantity is described completely by its size or magnitude.o A vector quantity has a magnitude (size or strength) and a direction.

The Components of a Vector (3.2)o Components add together to equal original vector.o Components must be perpendicular to each other.o The choice of components is a matter of convenience – which axes are better to

use for the particular problem.o Use trigonometry to resolve a vector into its components.

Vector Addition and Subtraction (3.3)o When adding (or subtracting) vectors, you must take into account both the

magnitude and the direction of the vectors.o Can add (subtract) vectors by two methods

Graphical method draw vectors head to tail lengths drawn to scale (use ruler) angles drawn accurately (use protractor)

Analytic method – add the components of the vectors and are the vector components and are perpendicular to each other.

Magnitude,

Direction: angle *** don’t forget to specify what the

angle, θ, is relative to (north, south, east, west, vertical, horizontal, etc) If , then and

Position, Displacement, Velocity, and Acceleration (in 2-D) (3.5)o Positiono Displacemento Average velocityo Instantaneous Velocityo Average accelerationo Instantaneous accelerationo These concepts that we first discussed for motion along a line (one dimension, x)

work just as well in a plane (two dimensions, x and y) Relative Velocity (3.6)

o Remember pattern of velocity subscriptso Review examples in book

vAB = vAG + vGB and vGB = -vBG

Motion in Two Dimensions (4.1)o Motions in x-direction and y-direction are independent of each other and occur

simultaneouslyx–component: y-component:

vx = v0x + axt vy = v0y + aytx = x0 + ½(v0x + vx)t y = y0 + ½(v0y + vy)tx = x0 + v0xt + ½ axt2 y = y0 + v0yt + ½ ayt2

vx2 = v0x

2 + 2axΔx vy2 = v0y

2 + 2ayΔy*** Make sure you understand the difference between initial velocity, v0, final velocity, vf, and average velocity, vave. The final velocity ONLY equals the average velocity when the acceleration is 0. ***

(Note: bold variables are vectors; otherwise, variables are scalars or magnitudes of vectors) Projectile Motion (4.2)

o Assume no air resistance.o In the x-direction (horizontal): moving object doesn’t slow down (change

velocity) if no air resistance. v0x = vx

ax = 0o In the y-direction (vertical): object is subject to gravity.

vy changes ay = g (acceleration due to gravity)

o Resulting motion is a curved trajectory (parabolic in shape)o Remember x and y motions are independent of each other. This means that a

package dropped from a hot air balloon at height H, will take exactly the same time to fall as a package dropped from an airplane traveling horizontally at height H. The y-components of the velocities are also the same for the two packages at all points on the way down.

o Speed of the projectile at any location along its path v = (vx2 + vy

2)1/2

horizontal and vertical components of velocity contribute to the speed or magnitude of velocity.

o At maximum height, vy = 0.o Zero launch angle (v0 is horizontal)

v0y = 0 x depends on v0x and time to fall (use y-component to get time)

o Range – horizontal distance before landing (y = y0)o Maximum range of projectile occurs when angle of initial velocity vector, q = 45°

(if there is no air resistance)o Projectiles have symmetries like freely falling bodies did.

Time for projectile to reach maximum height = time for projectile to fall from maximum height back to original height.

(General launch angle) Max height occurs at x = ½R v0x = vx (because ax = 0) magnitude of v0y = magnitude of vy when at same height (y0 = y)

- direction of v0y and vy NOT opposite- but θ above horizontal for v0 = θ below horizontal for v (when at same height)

The Concepts of Force and Mass (5.1)o Force is a vector quantityo Mass is a scalar quantity

Newton’s First Law of Motion (5.2)o An object stays at rest (velocity = 0), or in a constant state of motion at a constant

speed in a straight line, changed by a net force.o A net force applied to a body changes an object’s velocity – it doesn’t sustain it.o Inertia is the tendency of an object to remain in motion in a straight line (or at rest

where velocity = 0).o The inertia of an object is measured by its mass.

Newton’s Second Law of Motion (5.3)o F = mao As Components (5.5)

Fx = max

Fy = may

o F = 0 implies a = 0, and visa versa. Newton’s Third Law of Motion (5.4)

o “For every action, there is an equal, but opposite, reaction.”o FA on B = -FB on A

“Types of Forces” in mechanicso Common forces

Gravitational (Weight – at/near Earth’s surface) Normal Frictional forces Tension force Restoring force of a spring (Hooke’s Law)

Weight (5.6)o Weight = gravitational force that the earth (or another astronomical body like the

moon) exerts on an object Weight always acts downward, toward the center of the earth An object has weight regardless of whether or not it is resting on the

surface of the earth Weight depends on the acceleration at the surface of the Earth (or other

astronomical body like the moon), while the mass of an object does not change

The Normal Force (4.7)o The component of force that a surface exerts on an object with which it is in

contacto This component is perpendicular to the surfaceo The normal force is NOT always equal to the weight of the object lying on the

surface – it depends on what other forces are present.o Apparent weight (see textbook for discussion)

Static and Kinetic Frictional Forces (6.1)o Even highly polished surfaces are rough when looked at on a microscopic scaleo Static frictional force – doesn’t exist until you try to slide an object

fs £ fsmax (balances out force trying to slide an object to a certain point, a

maximum that occurs just before the object begins to slide) fs

max = msFn

ms = coefficient of static friction (typical values: 0.01 for smooth surfaces, and 1.5 for rough surfaces)

FN = magnitude of the normal force Direction – opposes sliding motion Independent of surface area

o Kinetic frictional force fk = mkFn

mk = coefficient of kinetic friction FN = magnitude of the normal force Direction – opposes sliding motion mk typically less than ms

The Tension Force (6.2 – Strings)

o A force is applied to right side of rope that is tied to a block (on the ropes left side) – every particle of the rope in turn applies a force to (pulls) its neighbor – finally in turn, the force is applied to the block. We say “the force T is applied to the block because of the tension in the rope.”

o Consider common usage of the word tension, where it is the tendency of the rope to be pulled apart. The right side of the rope in the situation above is pulled by force +F, while the left side of the rope is pulled the opposite direction (-F) by the block attached to the rope (Newton’s 3rd law)

o We have and will continue to use the concept of a “massless” rope – no net force is required to accelerate the rope, so the force applied to the rope is applied undiminished to the attached object

The Restoring force of Springs (6.2 – Springs)o Hooke’s law: FR = -kx

x = displacement of end of spring away from equilibrium position k = spring constant (also called the force constant) with SI units of N/m the minus sign means the force points in the opposite direction of the

displacement, x assume the spring itself has no mass (an ideal spring)

Translational Equilibrium (6.3)o F = 0

Fx = 0 and Fy = 0o An object in equilibrium is not accelerating, so there is no net force.o The net force on an object in equilibrium is zero, so the object is not accelerating.o Remember, an object with a constant velocity (is moving, but not changing

direction), is NOT being accelerated, so the net force is zero. Connected Objects (6.4)

o Two ways to approach problem Method #1: Treat each object separately, and

draw free body diagrams for each object separately adding up forces for each object separately remember that the acceleration is the same for all of the connected

objects (or the connecting string would stretch, break or bunch up) the magnitude of the tension is the same for all objects connected

by the same string Use algebra to combine separate equations and solve for the

unknowns. Method #2: Treat all the objects as a system

Mass of system = sum of individual masses Only consider forces external (outside) the system,

o so the tension in a string connecting objects in a system doesn’t count

remember that the acceleration is the same for all of the connected objects (or the connecting string would stretch, break or bunch up)

Circular Motion (6.5)o Motion of an object traveling at a constant (uniform)speed on a circular patho Direction is constantly changing, so the velocity is NOT constant

o Period, T – time required to travel once around the circle (one complete rotation)

o (speed)o Centripetal Acceleration

(magnitude)

Substituting for v, we get (magnitude) Direction – always points toward center of circle.

o Centripetal Force

(magnitude) Not a new force, just the name we give to the net force or component of

force that points toward the center of the circleo Banked Curves

Component of the normal force that points toward the center of the circle. (Occasionally, other forces or components of forces might contribute to the centripetal force in addition to the component of the normal force.)

Knowing the speed wanted and the radius of the (circular) curve, you can solve for the angle of the incline (bank). It is independent of the mass of the vehicle going around the curve. Example shown in class.

Work Done by a Constant Force (7.1)o W = (Fcosq)d

F = magnitude of force applied d= magnitude of displacement q = angle between the direction of the force applied and the direction of

the displacement (values between 0 and 180 degrees) Work is a scalar quantity, but can be either positive or negative SI units = Nm = J = Joule

The Work-Energy Theorem and Kinetic Energy (7.2)o Wnet = DKE = KEf - KE0 = ½ mvf

2 + ½ mv02

o Kinetic Energy: KE = ½ mv2

SI units = J = Joule Work done by a variable force (7.3)

o Work = area under the curve in a graph of Force versus Position (y vs x)o Work to stretch or compress a spring a distance x from equilibrium

WSp = ½ kx2

always positiveo Work done BY A SPRING

WSp = ½ kx2

May be POSITIVE or NEGATIVE depending on the situation Look at direction of force from the spring and the direction of the

displacement to see if the work is positive or negative (see also textbook, figure 7-12, Physics, by James S. Walker, 4th edition)

Power (7.4)

o

o For constant speed, P = Fv Conservative forces (8.1)

o Definition 1: “a force that does zero total work on any closed path.”o Definition 2: “If the work done by a force in going from arbitrary point A to an

arbitrary point B is independent of the path from A to B, the force is conservative.”

Gravitational force (Weight) Elastic Spring force Electric force (Phy 123)

Non-Conservative forces (8.1) Static and kinetic friction forces Air resistance Normal force Tension force

Potential Energy and Work Done by Conservative Forces (8.2)o Potential Energy, PEo Work done by conservative forces change the potential energy

WC = PE0-PEf = -(PEf-PE0) = -ΔPEo Gravity

PE = mg(h0 – hf) = mgho Springs

PE = ½ kx2

PE of spring is always positive or zero The Conservation of Mechanical Energy (8.3)

o Mechanical Energy, ME = PE + KEo If net work done by nonconservative forces is zero then mechanical energy is

conserved If WNC = 0, then Ef = E0

o This equation can also be written as½ mvf

2 + mghf + ½ kxf2 = ½ mv0

2 + mgh0 + ½ kx02

Work Done by Nonconservative forces (8.4)o If nonconservative forces do net work, then the mechanical energy changes

WNC = DME WNC = DKE + DPE WNC + KE0 +PE0 = KEf + PEf

How to do problems with work-energy1. Draw a picture2. Identify external conservative and non-conservative forces that act on an object.3. If the net work done by non-conservative forces is zero (WNC = 0), then Conservation

of Mechanical Energy applies (Ef = E0). Otherwise, the work done by non-conservative forces is equal to the change in mechanical energy (WNC = DKE + DPE).

4. Choose the location where PEgravity = 0 (where all heights in the problem are relative to). Don’t change this during the course of solving the problem. Usually PEspring = 0 at equilibrium position.

5. Now solve, using WNC = 0 Ef = E0, OR WNC = DKE + DPE, depending on what you found regarding the work done by external non-conservative forces.

The following is arranged for a different textbook, so it may seem out of order compared to Physics, 4th ed., by James S. Walker. – Dr. Nazareth, 2/17/2010.

The Impulse-Momentum Theorem (9.3)o Impulse = FDt = Dpo Linear Momentum = p = mv (9.1)

The Principle of Conservation of Linear Momentum (9.4)o If net external force is zero (Fext=0) then total momentum is conserved

p0 = pf

Collisions (9.5 and 9.6)o Elastic collision: KE0 = KEf

o Inelastic collision: KE0 KEf (energy before collision does NOT equal energy after)

o Completely inelastic collision: Objects collide and “stick together”, moving as one object after the collision with a combined mass.

It is vital that you understand whether a collision is ELASTIC or INELASTIC! Center of Mass (9.7)

o X-coordinate of the center of mass XCM = ∑(mx)/M

o Y-coordinate of the center of mass YCM = ∑(my)/M

o Velocity of the center of mass VCM = ∑(mv)/M

o Acceleration of the center of mass ACM = ∑(ma)/M

o Newton’s 2nd law for a System of particles MACM = ∑Fext = Fnet, external

Systems with Changing Mass – Rocket Propulsiono NOT REQUIRED, only for one or two extra credit homework problems in ch. 9

Angular Position (10.1)o θ (radians) = angle measured from reference line

counterclockwise rotation = + clockwise rotation = - 2π radians = 360 degrees = 1 revolution 1 rad. = 360 degrees/2π ≈ 57.3 degrees units for angles (radians or degrees or revolutions) = dimensionless – we

just write rad or radians to remind us which units we are using for angles Angular Displacement (10.1)

o Δθ (in radians) = θ –θ0 = s/r Note: I am often a bit lazy and assume θ0 = 0, so that Δθ = θ, and just use

θ for the angular displacement (not the position). A bad habit, I know … but so rarely is θ0 anything other than 0. – Dr. Nazareth

Angular Velocity (10.1)

o Average angular velocity Units = radians/second = rad./s = s-1

remember angles are dimensionless (see angular position) counterclockwise rotation = + clockwise rotation = -

o Instantaneous angular velocity, ω angular velocity at particular instant in time (in limit where Δt → 0) Units = radians/second = rad./s = s-1

usually just say angular velocity counterclockwise rotation = + clockwise rotation = -

o Angular speed = magnitude of angular velocity Angular Acceleration (10.1)

o Average angular acceleration Units = radians/second2 = rad./s2 = s-2

remember angles are dimensionless (see angular position) if Δω > 0 then α = + if Δω < 0 then α = - if ω and α have same sign then rotation is increasing if ω and α have opposite sign then rotation is decreasing

o Instantaneous angular acceleration, α angular velocity at particular instant in time (in limit where Δt → 0) Units = radians/second2 = rad./s2 = s-2

usually just say angular acceleration See average angular acceleration for sign convention

The Equations of Rotational Kinematics (10.2)o ω = ω0 + ato q = q0 + ½ (ω + ω0)to q = q0 + ω0t + ½ at2

o ω2 = ω02 + 2a(q - q0)

o Apply angular equations in the same way you did for the linear equations in chapter 2, just the physical interpretation is different.

Connection between Linear and Rotational Variables (10.3)o Tangential velocity vT = rω (vT in m/s, r in m and ω in rad/s)

o Centripetal acceleration (ac in m/s2, r in m, ω in rad/s) Due to changing direction of motion

o Tangential acceleration aT = ra (aT in m/s2, r in m and a in rad/s2) Due to changing angular speed If tangential speed is changing, then nonuniform circular motion.

o When both ac and aT are present

o (magnitude of the acceleration)

o (direction of the acceleration) Rolling Motion (10.4)

o Example of car tire rolling.o Linear speed of car = tangential speed of point on outer edge of the tire.

v = vT = rw (w in rad/s)o Linear acceleration of car = tangential acceleration of point on outer edge of the

tire. a = aT = ra (a in rad/s2)

Rotational Kinetic Energy and the Moment of Inertia (10.5)o I = moment of inertia (Table 10.1 lists for common objects)

Measure of the object’s resistance to a change in angular motion Depends on the mass of an object and how the mass is distributed about

the axis of rotation in question. For a system of particles I = ∑miri

2

SI units = kgm2

o Rotational Kinetic Energy: KER = ½ Iw2 (w in rad/s) Conservation of Energy (10.6)

o Total Kinetic Energy (rolling motion): KET = ½ mv2 + ½ Iw2

o Total Mechanical Energy: ME = ½ mv2 + ½ Iw2 + mgh + ½ kx2

ME = KEtranslational + KErotational + PEgravitational + PEspring

o Work-Energy Theorem: WNC + ½mv0

2 + ½Iw02 +mgh0 + ½kx0

2 = ½ mvf2 + ½ Iwf

2 + mghf + ½ kx2

If WNC = 0, then Ef = E0

Torque (11.1)o t = lF

Where: F = applied force (N), and l = lever arm (m). Lever arm = moment arm

o t = l x F (cross product of l and F; General definition of torque)o t = l(F sinφ) (force is at angle φ relative to radial line)o t = Fperpendicular d (d = distance force is applied from axis of rotation)o Convention: If force produces counterclockwise rotation = +t.

If produces clockwise rotation = -t. Torque and Angular Acceleration (11.2)

o t = Ia (I = moment of inertia; a = angular acceleration in rad/s2) o A net torque causes an angular acceleration

Magnitude depends on distribution of mass about axis of rotation/pivot point

Analogous to Newton’s 2nd law for translational motion Zero Torque and Static Equilibrium (11.3)

o Fx = 0 and Fy = 0 and t = 0 Center of Mass and Balance (11.4)

o Center of Gravity = where object’s weight can be considered to act when calculating the torque due to a weight = center of mass

Geometric center of regular shaped object with uniform mass distribution If object supported directly under center of gravity (center of mass), object

balances Angular Momentum (11.6)

o L = Iw (SI units = kg m2/s) I = moment of inertia w in rad/s)

o Angular momentum for a point particle L = rpsinφ φ = angle between p and radial direction

o Newton’s 2nd law for rotational motion t = Ia = ΔL/Δt

Conservation of Angular Momentum (11.7)o If text = 0 then Lf = L0.o Spinning figure skater example

Arms out = slower rotation because moment of inertia is larger (more mass farther from axis of rotation)

Arms in = faster rotation because moment of inertia is smaller (mass is closer to axis of rotation)

Rotational Work and Power (11.8) o Rotational Work: WR = t Δ q (q in rad) o Rotational Kinetic Energy: KER = ½ Iw2 (w in rad/s) o Total Kinetic Energy: KET = ½ mv2 + ½ Iw2

o Power produced by a torque: P = Work/Δt = t Δq/Δt = twave = tw if a = 0 Newton’s Law of Universal Gravitation (12.1)

o (magnitude of force)o m1 and m2 are mathematical pointso G = 6.67´10-11 Nm2/kg2 = universal gravitation constant – same value for all pairs

of particles everywhere in the universeo It is an attractive forceo The force that each particle exerts on the other is directed along the line joining

the particleso (12.2) Also true for spherical bodies (like planets or stars), but distance is to

center of the sphere (planet)o Weight = gravitational force that the earth (or another astronomical body like the

moon) exerts on an object

(Weight = mg is the approximation)

Weight always acts downward, toward the center of the earth An object has weight regardless of whether or not it is resting on the

surface of the earth Weight depends on G, the mass of the earth, ME, and distance, r, while the

mass of an object does not change Kelpler’s Laws of Orbital Motion (12.3)

o First law: Planets follow elliptical orbits with the Sun at one focus of the ellipse.o Second law: As a planet moves in its orbit, it sweeps out an equal amount of area

in an equal amount of time (See Figure 12-9 in textbook)o Third law: The period, T, of a planet increases as its mean distance form the Sun,

r, raised to the 3/2 power. That is, T = (constant)r3/2

For planet orbiting the Sun (circular orbit), T = (2π/sqrt(GMsun))r3/2

Same formula works for orbiting the Earth or another planet – just replace the mass of the Sun with the mass of the planet being orbitted

Note the period depends on the mass of the object being orbited, not the mass of the orbiting object

Satellites in Circular Orbitso Satellite is kept on circular path by gravitational force.

o (magnitude) → o If a satellite is to remain at radius, r, it must have speed, v, precisely.

o (magnitude) Gravitational Potential Energy (12.4)

o PEg = -GmM/ro PEg approaches zero as r approaches infinityo Convenient to choose infinity as reference point for PEg = 0 when have distances

on astronomical scale Energy Conservation (12.5)

o Mechanical energy of an object of mass, m, at distance, r, from the Earth ME = PE + KE = ½ mv2 – GmME/r

o Speed of asteroid striking earth, starting at infinite distance with v0 = 0 vf = sqrt(2GME/RE) = 11,200 m/s ≈ 25000 miles per hour

o Escape speed of object just barely leaving Earth, with vf = 0 when reaches infinite distance from Earth.

This is the initial velocity it must have at the surface to escape Earth’s gravity

v0 = sqrt(2GME/RE) = 11,200 m/s ≈ 25000 miles per hour