Unit 21 – Session 1 Last semester, we explored a number of different force types:

1) Gravitational force, .

2) Normal force,

3) Tension force,

4) Friction force,

It turns out that the last three forces are all just due to the electric force at the molecular level – charges repelling each other (normal) or attracting each other (tension, friction)

!Fgrav = m !g

!Fnorm

!Ftens

!F fric

It also turns out that two objects with the property of mass interact with each other in an extremely similar way as two objects with the property of charge. In fact, there is a gravitational force law (Newton’s universal law of gravitation) for two objects with mass that looks very similar to Coulomb’s law for two objects with excess charge.

Electrical

(Coulomb’s law) Gravitational

(Newton’s universal law of gravitation)

The expression we’ve previously been using, , is a special case for when the Earth is one of the two interacting objects with mass, and the other object is near the surface of the Earth.

!FonAbyB

elec = kqAqBr2

r̂!FonAbyB

grav = –GmAmB

r2r̂

k = 8.99 ×109 N ⋅m2

C2G = 6.67 ×10−11 N ⋅m2

kg2

!Fgrav = m !g

Activity 21.2.1. Be sure to try answering parts a., b., and c. before moving on to the next page of these notes. a. Try looking at each individual term in the two expressions to see

how they are similar to each other (e.g., how are k and G similar in concept?).

b. Try looking at each individual term in the two expressions, to see how they are different from each other.

c. Keep in mind that nature is, for the most part, pretty consistent – how do two positive masses interact with each? How do two positive charges interact with each other? Two negative charges?

Activity 21.2.1. (continued) Are you sure you are ready to move on to the next page? Did you really dig into trying to find the similarities and differences in the two force laws?

Activity 21.2.1. (continued) Hopefully, you came up with these ideas. a. Similarities:

1) (k) and (G) — both force laws depend on a universal constant (“universal” means the constant has the same value everywhere in the universe, not just here on Earth).

2) (qAqB) and (mAmB) — both force laws depend on the product of a property of the two objects.

3) — both force laws are inverse square laws (the magnitude of each force decreases a lot as the objects are separated more and more).

4) — both forces are vectors, with the forces acting parallel to a line connecting the centers of the two objects (the force on one object is directly toward or directly away from the other object).

1 r2

r̂

Activity 21.2.1. (continued) b. Differences:

1) (k) and (G) — In addition to units (C2 vs. kg2), the values are significantly different, with k being very large and G being very small.

2) (qAqB) and (mAmB) — depends on the property of charge, while depends on the property of mass. Also, there are two types of charge, + and –, but only one type of mass, +.

3)

4) — is repulsive for like charges, and attractive for unlike charges (this is taken care of by the + and – signs of the charges).

is always attractive (this is taken care of by the – sign in the expression for the gravitational law, making always in the opposite direction of ).

!Felec

!Fgrav

r̂!Felec

!Fgrav

!Fgrav

r̂

Activity 21.3.1

● Part c. – Just to make sure your calculations in parts a. and b. are working ok, you

should get: .

● Part d. – There is no one answer to this question – it is whatever your experience is. For example, when I do an endo when mountain biking, I’m very aware of the gravitational force. On the other hand, just combing my hair makes it out of control due to the electrical force from the excess charge. In a more subtle way, sitting in this chair for 4 hours of class makes my butt sore due to the normal force – which is ultimately an electrical force – balancing the gravitational force.

factor =

!Felec

!Fgrav = 2.3×1039

Activity 21.4.1

● Revisit your answer to Activity 20.5.2 on page 562. Remember that the electrical Gauss’ law is , where the electric field vector was defined

as .

● Look back at Activity 21.2.1 to remind yourself of the similarities and differences between Coulomb’s law and Newton’s universal law of gravitation, and to use them to help you make the connection between electrical Gauss’ law and gravitational Gauss’ law. (Don’t forget that Newton’s universal law of gravitation has a minus sign – it’s important here as well.)

!E!A cosθ = 4π kqenclosed

!E ≡!Felec q

Activity 21.4.2 ● Part a. – Here, you will draw a spherical Gaussian surface around the Earth at a

distance h just above the surface, very similar to the Gaussian surface you drew just inside the can surface in Activity 20.6. Remember that the radius, r, of your sphere is measured from the center of the sphere (which here is also the center of the Earth). The surface area of a sphere is .

● Part b. – Keep in mind that the gravitational field vector, , points toward the center of the Earth, and the area vector, , of the Gaussian surface points away from the center of the Earth, so .

● Part d. – Again, remember that the radius, r, is measured from the center of the Gaussian sphere –

● Part f. – Even though the gravitational field, , in part d. didn’t seem to be that large, remember that the gravitational force on an object in a gravitational field is given by . For an Apollo space capsule, this would be

. What gives us our sensation of weight that is missing for the astronauts? As shown below, they have 96% of their Earth weight.

At 120 km, (96% of )

!A = 4πr2

!Y

!A

θ = 180°

r = 384,000km 2 = 1.92 ×108m!Y

!Fgrav = m

!Y

!Fgrav = 28,800kg( ) 0.011N kg( ) = 320N

!Y =

− 6.67 ×10−11 N ⋅m2 kg2( ) 5.98 ×1024 kg( )6.38 ×106m +1.20 ×105m( )2 cos180°

= 9.4 Nkg

!g

Activity 21.5.1 ● Part a. – The image on the next page of these notes shows the reading on the 5.0 N

spring scale as I pulled the cart (at constant speed) along the inclined plane. I pulled the bottom left of the cart from point a to point c. Be careful here; when calculating the work, the angle, , is not 30°. Which direction does point? Which direction does point? (You did something similar last semester when you bid on the 3 jobs lifting boxes in Activities 10.2.1 and 10.4.2.)

● Part b. – The image following the one on the next page shows the reading on the 5.0 N spring scale as I lifted the cart from point a to point d, then moved it sideways from point d to point c, all at constant speed. Again, for each of the two separate paths, think about the angle, , between the vectors and .

θ!Fspring scale

Δ !s

θ!Fspring scale Δ !s

Activity 21.5.1 a.

19.8

4 cm

30°

Fd c

a b

39.68 cm

Activity 21.5.1 b.

F

Activity 21.6.1 For all three cases, both the electric field, , and the angle, , are constant for the whole path, so we can use:

(no scary integral needed)

Unless otherwise stated, it is usually assumed that a test charge, qt, is a positive type of charge.

!E θ

W elec =!Felec Δ !s cosθ

= qt!E Δ !s cosθ

cos0° = 1cos45° = 0.71cos90° = 0cos135° = −0.71cos180° = −1