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Particle Mass Electric Charge

Electron me = 9.110 x 10-31 kg q = -e q = -1.60 x 10-19 C

Proton mp = 1.673 x 10-27 kg q = +e q = +1.60 x 10-19 C

Neutron mn = 1.675 x 10-27 kg q = 0

q = 0 C

Electrostatics 1) electric charge: 2 types of electric charge: positive and negative 2) charging by friction: transfer of electrons from one object to another 3) positive object: lack of electrons negative object: excess of electrons

Conservation of Electric Charge: The total electric charge of an isolated system remains constant.

4) Types of materials:

a) Conductors: materials in which electric charges move freely (e.g. metals, graphite) b) Insulators: materials in which electric charges do not move freely (e.g. plastic, rubber, dry wood, glass, ceramic) c) Semiconductors: materials with electrical properties between those of conductors and insulators (e.g. silicon) d) Superconductors: materials in which electrical charges move without resistance (e.g. some ceramics at very low temperatures)

Properties of Atomic Particles

e = elementary unit of charge (magnitude of charge on electron) e = 1.60 x 10-19 C

1. A balloon has gained 2500 electrons after being rubbed with wool. What is the charge on the balloon? What is the charge on the wool?

2. A rubber rod acquires a charge of -4.5 µC. How many excess electrons does this represent?

q = -4.0 x 10-16 C q = +4.0 x 10-16 C

2.8125 x 1013 e

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Electric Force (Electrostatic Force, Coulomb Force)

Coulomb’s Law: The electric force between two point charges is directly proportional to the product of the two charges and inversely proportional to square of the distance between them, and directed along the line joining the two charges.

NOTE: +-F denotes direction of force not sign of charge

1 22

0

1 22

0

14

4

e

e

q qr

q qFr

Fπε

πε

=

=

1 22eF

q qkr

=k = Coulomb constant (electrostatic constant) k = 8.99 x 109 N·m2 C -2

k = 1/ 4πε0 ε0 = permittivity of free space = 8.85 x 10-12 C2 N-1 m -2

Coulomb Force

Point charge: a charged object that acts as if all its charge is concentrated at a single point

Alternate formula for Coulomb force:

Use the Coulomb force to estimate the speed of the electron in a hydrogen atom.

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The Principle of Superposition

The net electric force acting on a charged particle is the vector sum of all the electric forces acting on it.

1. Determine the net electrostatic force on charge q1, as shown below.

2. Where can a third charge of +1.0 µC be placed so that the net force acting on it is zero?

3. Three point charges of -2.0 µC are arranged as shown. Determine the magnitude and direction of the net force on charge q1.

D = 2/3 m

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Electric Field

Electric field: a region in space surrounding a charged object in which a second charged object experiences an electric force

Test charge: a small positive charge used to test an electric field

Electric Field Diagrams 1. Positively charged sphere 2. Positive point charge 3. Negative point charge

Radial Field: field lines are extensions of radii

5. Two positive charges 6. Two negative charges 7. Two unlike charges

Properties of Electric Field Lines 1. Never cross 2. Show the direction of force on a small

positive test charge 3. Out of positive, into negative 4. Direction of electric field is tangent to

the field lines 5. Density of field lines is proportional

to field strength (density = intensity) 6. Perpendicular to surface 7. Most intense near sharp points

8. Oppositely charged parallel plates

Edge Effect: bowing of field lines at edges

Uniform Field: field has same intensity at all spots

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Electric Field Strength (Intensity): electric force exerted per unit charge on a small positive test charge

Electric Field Strength

Units: N/C

eFEq

= 2

2 20

14

Qqk Q QrE kq r rπε

= = =

Electric Field: Electric Field for a Point Charge:

Point Charge Spherical Conductor

1. a) Find the magnitude and direction of the electric field at a spot 0.028 meter away from a sphere whose charge is +3.54 microcoulombs and whose radius is 0.60 centimeters.

b) Find the magnitude and direction of the electric force acting on a -7.02 nC charge placed at this spot.

c) Find the electric field strength at the surface of the sphere.

2. a) Find the magnitude and direction of the gravitational field at an altitude of 100 km above the surface of the Earth.

b) Find the magnitude and direction of the gravitational force exerted on a 6.0 kg bowling ball placed at this spot.

c) Find the gravitational field strength at the surface of the Earth.

Electric Force:

eF Eq=Units: N

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3. a) Find the magnitude and direction of the net electric field halfway between the two charges shown below.

b) Determine the electric force on a proton placed at this spot.

4. Two charged objects, A and B, each contribute as follows to the net electric field at point P: EA = 3.00 N/C directed to the right, and EB = 2.00 N/C directed

downward. What is the net electric field at P?

E = 3.61 N/C Theta =33.70

5. a) Two positive point charges, q1 = +16 µC and q2 = +4.0 µC, are separated in a vacuum by a distance of 3.0 m.

Find the spot on the line between the charges where the net electric field is zero.

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6. A proton is released from rest near the positive plate. The distance between the plates is 3.0 mm and the strength of the electric field is 4.0 x 103 N/C.

a) Describe the motion of the proton. constant acceleration in a straight line b) Write an expression for the acceleration of the proton.

c) Find the time it takes the proton to reach the negative plate. d) Find the speed of the proton when it reaches the negative plate.

7. A particle is shot with an initial speed through the two parallel plates as shown. a) Sketch and describe the path it will take if it is a proton, an electron, or a neutron. b) Which particle will experience a greater force? c) Which particle will experience a greater acceleration? d) Which particle will experience a greater displacement?

8. In the figure, an electron enters the lower left side of a parallel plate capacitor and exits at the upper right side. The initial speed of the electron is 5.50×106 m/s. The plates are 3.50 cm long and are separated by 0.450 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

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Electric Potential Energy

Gravitational Potential Energy (EP)

Base level where EP = 0

High amount of EP

Low amount of EP

Reason for EP: 1. Test object has mass (test mass = m) 2. Test mass is in a gravitational field (g) caused by larger object (M) 3. Larger object exerts a gravitational force on test mass (Fg = mg) 4. Test mass has tendency to move to base level due to force 5. Work done moving object between two positions is path independent. Gravitational potential energy: EP = mgh W = ΔEP = mg Δh

Electric Potential Energy (EP)

High amount of EP

Low amount of EP

Reason for EP: 1. Test object has charge (test charge = +q) 2. Test charge is in an electric field caused by larger object (Q) 3. Larger object exerts an electric force on test charge (FE = Eq) 4. Test charge has tendency to move to base level due to force 5. Work done moving object between two positions is path independent. Electric potential energy: EP = Eq h W = ΔEP = Eq Δh

Base level where EP = 0

Electric Potential Energy (EP)- the work done in bringing a small positive test charge in from infinity to that point in the electric field

EP = 0

2

cos

( )

P

P

r

P

r

P

P

E W FsE Eq s

kQE q dss

kQqEskQqEr

θ

∞

∞

= =

= − Δ

= −

=

=

∫

(Work done by field) Derivation for Point Charges

Electric Potential Energy due to a point charge

PkQqEr

=

Formula: Units:

J

Type: scalar

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Electric Potential (V) - work done per unit charge moving a small positive test charge in from infinity to a point in an electric field.

Electric Potential due to a point charge

04

P

kQqE rVq qkQ QVr rπε

= =

= =

Formula: Units:

J/C = volts(V)

Type: scalar

A B Higher potential Lower potential

Zero potential

A B

Lower potential Higher potential Zero potential

PE qV=

1. a) Calculate the potential at a point 2.50 cm away from a +4.8 µC charge.

b) How much potential energy will an electron have if it is at this spot?

3. What is the potential where a proton is placed 0.96 m from a -1.2 nC charge?

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Point Charges

+Q +Q +Q +Q

Electric Force Electric Field Electric Potential Energy Electric Potential

Two objects needed – interaction between the two Magnitude: F = Eq F = kQq/r2

Units: N Type: vector Direction: likes repel, unlikes attract Sign: don’t use when calculating – check frame of reference

One object needed – property of the field Magnitude: V = EP/q V = kQ/r Units: J/C Type: scalar (+/-) Sign: use sign of Q

Two objects needed – quantity possessed by the system Magnitude: EP = qV EP = kQq/r Units: J Type: scalar (+/-) Sign: use signs of Q and q

One object needed – property of that one object Magnitude: E = F/q E = kQ/r2 Units: N/C Type: vector Direction: away from positive, towards negative Sign: don’t use when calculating – check frame of reference

F = 0 where E = 0 EP = 0 where V = 0

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1. a) Calculate the net electric field at each spot (A and B):

b) Calculate the net electric force on a proton placed at each spot.

2. a) Calculate the net electric potential at each spot (A and B):

b) Calculate the electric potential energy of a proton placed at each spot.

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Electric Potential and Conductors

For a hollow or solid conductor, 1. all the charge resides on the outside

surface 2. the electric field is zero everywhere within 3. the external electric field acts as if all the

charge is concentrated at the center 4. the electric potential is constant (≠ 0)

everywhere within and equal to the potential at the surface

Distance

Elec

tric

Fiel

d St

reng

th Value at surface = kQ/r2

radius

Graphs for a spherical conductor

Distance

Elec

tric

Pote

ntia

l

radius

A spherical conducting surface whose radius is 0.75 m has a net charge of +4.8 µC. a) What is the electric field at the center of the sphere? b) What is the electric field at the surface of the sphere? c) What is the electric field at a distance of 0.75 m from the surface of the sphere? d) What is the electric potential at the surface of the sphere? e) What is the electric potential at the center of the sphere? f) What is the electric potential at a distance of 0.75 m from the surface of the sphere?

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Equipotential Surfaces

Equipotential surface: a surface on which the electric potential is the same everywhere

1. Locate points that are at the same electric potential around each of the point charges shown.

2. Sketch in the electric field lines

for each point charge. 3. What is the relationship between

the electric field lines and the equipotential surfaces?

Perpendicular

Field lines point in direction of decreasing potential

Electric Potential Gradient The electric field strength is the negative of the electric potential gradient.

Formula: VE

xΔ

= −Δ

Units: N/c or V/m

For each electric field shown, sketch in equipotential surfaces.

Sketch in equipotential surfaces for the two configurations of point charges below.

http://www.surendranath.org/Applets.html http://wps.aw.com/aw_young_physics_11/0,8076,898593-,00.html

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Electric Potential Difference (ΔV) – work done per unit charge moving a small positive test charge between two points in an electric field

P

WVqEVq

Δ =

ΔΔ =

Electric Potential Difference

Formula:

Units: J/C = V

High and Low Potential

1. a) Which plate is at a higher electric potential? positive b) Which plate is at a lower electric potential? negative c) What is the electric potential of each plate? Arbitrary – relative to base level d) What is the potential difference between the plates? Not arbitrary – depends on charge, distance between, strength of electric field, geometry of plates, etc. Mark plates with example potentials, as well as

spots within field Mark “ground” – mark equipotentials e) Where will: a proton have the most electric potential energy?

an electron? a neutron? an alpha particle? Not arbitrary

2. An electron is released from rest near the negative plate and allowed to accelerate until it hits the positive plate. The distance between the plates is 2.00 cm and the potential difference between them is 100. volts.

Eo = Ef Ee = EK

qV = ½ mv2 v = sqrt (2qV/m)

v = sqrt(2(1.6 x 10-19)(100 V) /(9.11 x 10-31))

v = 5.9 x 106 m/s

b) Calculate the strength of the electric field.

a) Calculate how fast the electron strikes the positive plate.

Formula: qV = ½ mv2 Ve = ½ mv2

Formula: E = -ΔV/Δx E = V/d

PE q VΔ = Δ

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4. In Rutherford’s famous scattering experiments (which led to the planetary model of the atom), alpha particles were fired toward a gold nucleus with charge +79e. An alpha particle, initially very far from the gold nucleus, is fired at 2.00 × 107 m/s directly toward the gold nucleus. Assume the gold nucleus remains stationary. How close does the alpha particle get to the gold nucleus before turning around? (the “distance of closest approach”)

2.74 x 10-14 m

The Electronvolt

Electronvolt: energy gained by an electron moving through a potential difference of one volt

ΔEe = qΔV ΔEe = (1e)(1 V) = 1 eV

ΔEe = (1.6 x 10-19 C)( 1 V) ΔEe = 1.6 x 10-19 J Therefore: 1 eV = 1.60 x 10-19 J

Derivation:

1. How much energy is gained by a proton moving through a potential difference of 150. V?

150 eV or 150(1.60 x 10-19) = 2.4 x 10-17 J

2. A charged particle has 5.4 x 10-16 J of energy. How many electronvolts of energy is this?

Factor-label (5.4 x 10-16 J) (1 eV/1.6 x 10-19) = 3375 eV

3. An electron gains 200 eV accelerating from rest in a uniform electric field of 150 N/C. Calculate the final speed of the electron.

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Universal Gravitation Kepler’s Three Laws of Planetary Motion

Law 1: All planets orbit the Sun in elliptical paths with the Sun at one focus.

Law 2: An imaginary line joining any planet to the Sun sweeps out equal areas in equal time intervals.

Law 3: The square of the orbital period of any planet is proportional to the cube of its average orbital radius.

Formula

2 3

2 3

32

T rT kr

T r

α

α

=

Newton’s Law of Universal Gravitation

Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses and that is inversely

proportional to the square of the distance between them.

Two approximations used in deriving the law:

1. Masses are considered to be point masses. Point mass: infinitely small object (radius = 0) whose mass is m 2. The force between two spherical masses whose separation is large compared to their radii is the same as if

the two spheres were point masses with their masses concentrated at the centers of the spheres.

Sun

Mean radius = 6.96 x 108 m

Earth

Mean radius = 6.37 x 106 m

Mean Earth-Sun distance = 1.50 x 1011 m

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2

2

2

2

2

2 2

2

2 2

3

2but

2

4

4

c cF maGMm mvr rGM vr

rvT

GM rr TGM rr TTr GM

π

π

π

π

Σ =

=

=

=

# $= % &' (

=

=

( )( )( )( )

( ) ( )

2 2

3

22

3 1111

30

30

4

1 365.25 24 3600 46.67 101.50 10

2.0 10

accepted value=1.99 10

s

S

Tr GM

year days hours s

x Mx m

M x kg

x kg

π

π−

=

# $% & =

=

Newton’s Derivation of Kepler’s Third Law What provides the centripetal force for orbital motion? gravitation

Derivation Application – “weighing the Sun”

Formula

1 22

1 22

=

g

g

m mFr

G m mFr

α⋅

⋅ ⋅

Point mass Extended spherical body

Re 2Re 3Re 4Re Gravitational Constant : G = 6.77 x 10-11 N m2/kg2

What is the resultant gravitational force on the Earth from the Sun and Moon, as shown below?

Average Earth-Sun distance = 1.50 x 1011 m

Average Earth-Moon distance = 3.84 x 108 m

Earth Mass = 5.98 x 1024 kg

Moon Mass = 7.36 x 1022 kg

Sun Mass = 1.99 x 1030 kg

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Gravitational Field Strength

Gravitational field strength at a point in a gravitational field:

the gravitational force exerted per unit mass on a small/point mass

Symbol: g Formula: g = Fg / m Units: N/kg (m/s2) Type: vector

Deriving formula for gravitational field strength at any point above the

surface of a planet

g = Fg /m

g = (GMm/r2)/m

g = GM/r2

Deriving formula for gravitational field strength at the surface of a planet

g = GM/r2

go = G Mp / Rp2

1. What is the gravitational field strength of the Earth at its surface?

“g” at the surface of the Earth

go = G ME/Re2

2. What is the gravitational field strength at an altitude equal to the radius of Earth?

“g” ratio

g/go = RE2/r2

Point mass

Extended spherical body

Re 2Re 3Re 4Re

go

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3. a) What is the resultant gravitational field strength at a point midway between the Earth and Moon?

Average Earth-Moon distance = 3.84 x 108 m

Earth Mass = 5.98 x 1024 kg

Moon Mass = 7.36 x 1022 kg

4. a) Is there a point where the resultant gravitational field strength of the Earth and Moon is zero? If so, where?

b) What is the resultant gravitational force acting on a 1500. kg space probe at this location?

b) What is the resultant gravitational force acting on a 1500. kg space probe at this location?

Average Earth-Moon distance = 3.84 x 108 m

Earth Mass = 5.98 x 1024 kg

Moon Mass = 7.36 x 1022 kg

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Gravitational Potential Energy

Difference in gravitational potential energy between any two points in a gravitational field: ΔEP = mgΔh

This difference is path independent. 1. Same ΔEP between any two points no matter what path is

taken between them. 2. Work done in moving a mass between two points in a

gravitational field is independent of the path taken. 3. ΔEP is zero between any two points at the same level no matter

what path is taken. 4. ΔEP is zero for any closed path (a path that begins and ends at

same point). “Old” formula for gravitational potential energy: Ep = mgh

discuss 2 problems with definition 1) “g” varies above surface 2) arbitrary base level

Gravitational Potential Energy of a mass at a point in a gravitational field:

the work done in bringing a small point mass in from infinity to that point in the gravitational field

Base level: infinity Gravitational PE at infinity: zero

EP = 0 EP = -100 J EP = -400 J

Derivation of gravitational potential energy formula

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Ep = - GMm/r EP = -Gm1m2/r

Formula not valid inside planet

Formula:

Units: J

Type: scalar

Potential energy vs. distance:

What is the gravitational potential energy of a 5000 kg satellite:

b) orbiting the Earth at an altitude of 200 km?

a) on the surface of the Earth?

Gravitational Potential

Gravitational potential at a point in a gravitational field:

work done per unit mass to bring a small point mass in from infinity to that point in the gravitational field

V = Ep/m

V = -GM/r

ΔV = W/m

ΔV = ΔEp/m

Difference in gravitational potential:

Gravitational potential at a point:

Ep at surface Ep = -GMpm/Rp

Units: J/kg

Type: scalar

Gravitational Potential vs. distance

Formulas:

Symbol: V V at surface Vo = -GMp/Rp

Symbol: V

c) How much does the potential energy of the satellite increase when it is put into this orbit?

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1. What is the gravitational potential due to the Earth’s gravitational field:

a) at the surface of the Earth? b) At a location three Earth radii from the center of the Earth?

c) What is the change in potential in moving from the surface to this new location?

d) What is the minimum amount of energy needed to lift a 5000 kg satellite to this location?

2. What is the net gravitational potential at a spot midway between the Earth and the Sun?

Sun Earth

3. Derive an expression for the gravitational potential at the surface of a planet in terms of the gravitational field strength.

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Escape Speed

Escape Speed: minimum initial speed an object must have at the surface of a planet in order to escape the gravitational attraction of the planet

Planet

Travel to infinity Just make it – means velocity is zero at infinity – means EK is zero at infinity as well as EP

Eo = Ep + Ek Ef = Ep

Assumptions: planet is isolated – ignore air resistance

Derivation:

Note: 1. Direction of travel is irrelevant – ΔEp is path independent 2. independent of mass of rocket 3. More speed (EK) is needed in real life since air friction is not negligible at lower altitudes

1. What is the escape speed for Earth? 2. If the Earth became a black hole, how large would it be?

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Satellite Motion

Natural Satellites Artificial Satellites

Acceleration of a satellite:

ac = v2/r

ac = g = GM/r2

ac = 4π2 r/T2

v = 2π r/T

Orbital speed of a satellite:

2

2

2

c cF maGMm mvr rGM vrGMvr

Σ =

=

=

=

v = 2π r/T

2. What happens to the required orbital speed if: a) the mass of the satellite increases?

Nothing – speed is independent of mass

1. Compare the motion of satellites A and B. A – faster, less time (smaller period) B – slower, more time (longer period) T2/R3 = constant true for all satellites T = kR3/2

b) the satellite is boosted into a higher orbit? Satellite would orbit at a slower

(tangential) speed

Period of a satellite:

2 3

2 3

32

T rT kr

T r

α

α

=

3. What would happen to a satellite if it encountered appreciable air resistance? Slow down, drop to lower orbit, and speed up, encounter even more air molecules (denser), cycle continues – spiral to Earth

“Weightlessness” Free fall Orbital motion Deep space

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Energy of Orbiting Satellites

Gravitational Potential Energy Kinetic Energy Total Energy

Compare the energies of the two orbiting satellites.

Energy Derivations

Gravitational Potential Energy

Kinetic Energy Total Energy

Graphs of the energies of an orbiting satellite

Comparisons:

RE

A 1500 kg satellite is to be put into orbit around the Earth at an altitude of 200 km.

a) How much potential energy will the satellite have at this altitude?

b) How much kinetic energy will the satellite need to orbit at this altitude?

c) What is the total amount of energy the satellite has at this altitude?

d) What is the orbital speed of the satellite?

e) What is the minimum amount of energy needed to lift the satellite from the surface of the Earth to this altitude?

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Comparisons

Equipotential surface: a surface on which the potential is the same everywhere

one point mass two point masses

1. The gravitational force does no work as a mass moves on along equipotential surface.

2. The work done in moving a

mass between equipotential surfaces is path independent.

3. The work done in moving a

mass along a closed path is zero.

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On the diagram at right: a) Sketch the gravitational field around the point mass. b) Sketch equipotential surfaces around the point mass.

What is the relationship between the gravitational field and the equipotential surfaces?

Perpendicular Field lines point in direction

of decreasing potential

Gravitational Potential Gradient

derive g = -ΔV/Δr

gravitational potential gradient: the gravitational field is the negative gradient of the gravitational potential with respect to distance

gradient: rate of change with respect to something “slope” or “derivative”

Formula -80 J/kg

-70 J/kg A

B

What is the average gravitational field strength between equipotential surfaces A and B if they are 5.0 m apart?

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Practice Questions

1. a) Calculate the gravitational force the Sun exerts on the Earth.

b) Compare this to the gravitational force that the Earth exerts on the Sun.

2. a) Calculate the strength of the gravitational field of the Sun at a location one million kilometers from the Sun.

b) What is the Sun’s gravitational force at this point?

3. a) Calculate the strength of the Sun’s gravitational field at the surface of the Earth.

b) Explain why the net gravitational field strength at the surface of the Earth can be approximated as due solely to the Earth’s gravitational field.

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4. a) Calculate the resultant gravitational field at a spot midway between the Earth and Sun.

b) Compare the contributions from the Sun and the

Earth to this resultant field. c) What is the gravitational force acting on a 5000 kg

space probe at this location?

Mars

5. A 5000kg satellite orbits Mars at a distance of 1000 km.

e) What is the minimum energy needed to lift the satellite to this altitude?

d) How much gravitational potential energy does the satellite have at this altitude?

c) What is the gravitational potential at orbiting altitude?

b) How much gravitational potential energy does the satellite have on the surface of Mars?

a) What is the gravitational potential at the surface of Mars?

Mass of Mars: 6.42 x 1023 kg Mean planetary radius: 3.37 x 106 m

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6. A 5000. kg satellite is placed in a low altitude orbit.

a) If the altitude is sufficiently low, what is the approximate radius of the satellite’s orbit?

b) Calculate the satellite’s orbital speed.

c) Calculate the orbital period of the satellite.

d) Calculate the gravitational potential energy of the satellite.

e) Calculate the kinetic energy of the satellite.

f) Calculate the total energy of the satellite.

g) What is the minimum amount of energy needed to lift the satellite into this orbit?