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Physics 1710—Warm-up Quiz

Answer Now !

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35% 50 of 140

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In the movie “Armaggedon” a comet “the size of Texas’” (800 miles or 1250 km in diameter) was heading for Earth. If the density were 800 kg/m3, the mass would be about 8.x10 20 kg. How far would a ball on the surface fall in 1.0 second?

  • About 5 m
  • About 7 cm
  • About 2 cm
  • None of the above
gravitation

Physics 1710—Chapter 12 Apps: Gravity

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g =G M/ R2

g=(6.67x10-11 N m2/kg2)(8x1020 kg)/(625x103 m)2

g = 0.14 N/kg

d = ½ g t 2 = 0.5 (0.14 m/s2)(1 sec)2

= 0.0 7 m = 7 cm

Gravitation

slide3

Physics 1710—Chapter 13 Apps: Gravity

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1′ Lecture

The gravitational force constant g is equal to g = G M/(R+h) 2, M and R are the mass and radius of the planet.

The gravitation field is the force divided by the mass.

The gravitation potential energy for a point mass is proportional to the product of the masses and inversely proportional to the distance between their centers.

The escape velocity is the initial velocity at the surface required to leave a body.vescape = √[ 2GM/R]

does the position of the moon affects us as astrology claims

Physics 1710—Chapter 13 Apps: Gravity

No Talking!

Think!

Confer!

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Does the position of the moon affects us as astrology claims?

Peer Instruction Time

does the position of the moon affects us as astrology claims5

Physics 1710—Chapter 13 Apps: Gravity

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34% 48 of 140

Answer Now !

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Does the position of the moon affects us as astrology claims?
  • Yes, I strongly agree
  • Yes, I Agree
  • No, I disagree
  • No, I strongly disagree
astrology does the moon affect us f g m m d 2

Physics 1710—Chapter 13 Apps: Gravity

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M = 7.36x1023 kg; d = 3.84x108 m

|F/m|= (7x10-11)(7x1023)/(4x108)2 = 3x10-4 N/kg ~ 30 μ g

Astrology: Does the moon affect us?

F = - G M m/ d 2

d moon=3.84x108 m

M =7.36x1023 kg

gravitational potential energy u r f d r u r g mm r 2 d r u gmm r mgr u 0 as r

Physics 1710—Chapter 13 Apps: Gravity

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Gravitational Potential Energy

U = -∫∞RF•d r

U = -∫∞R G Mm/r 2d r

U = GMm/R = mgR

U = 0 as r →∞

F

d r

escape velocity if k is such that e 0 as r then k r m v 2 u r gmm r and v escape 2gm r 2gr

Physics 1710—Chapter 13 Apps: Gravity

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Escape Velocity

If K is such that E = 0 as r →∞, then

KR = ½ m v 2 = UR = GMm/ R

and

vescape = √[ 2GM/R] = √[2gR]

UR = GMm/R

U∞ = 0

K∞ = 0

K = ½ m v 2

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Physics 1710—Chapter 13 Apps: Gravity

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37% 52 of 140

Answer Now !

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Were the astronauts in danger of jumping off the “Armaggedon” comet? (Hint: What is the escape velocity from our “Armaggedon” comet? M = 8x1020 kg, R =6.25x10 5 m, g = 0.14 N/kg)

  • Yes. Definitely.
  • Maybe.
  • No. Definitely not.
  • Not enough information.
escape velocity v escape 2gm r 2gr 2 0 14 n kg 625 km 400 m s 900 mph safe

Physics 1710—Chapter 13 Apps: Gravity

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Escape Velocity

vescape = √[ 2GM/R] = √[2gR]

= √[2(0.14 N/kg)(625 km)] ~ 400 m/s

~ 900 mph. Safe!

vescape = √[ 2GM/R] = √[2gR]??

escape velocity of earth v escape 2gm r 2gr 2 9 8 n kg 6 3 x10 6 m 11 000 m s 25 000 mph stuck

Physics 1710—Chapter 13 Apps: Gravity

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Escape Velocity of Earth?

vescape = √[ 2GM/R] = √[2gR]

= √[2(9.8 N/kg)(6.3 x10 6 m)] ~ 11,000 m/s

25,000 mph. Stuck!

vescape = √[ 2GM/R] = √[2gR]??

slide12

Physics 1710—Chapter 13 Apps: Gravity

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Total Energy

for Gravitationally Bound Mass

E = K + U

E = ½ m v 2 – GMm/r

E = L2/2mr 2 – GMm/r

Bound orbit if E ≤ 0 and

- dE/dr |ro= 2 (L2/2mro) - GMm = 0

E = - GMm/2ro

total energy for gravitationally bound mass e k u e gmm 2r o

Physics 1710—Chapter 13 Apps: Gravity

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Total Energy

for Gravitationally Bound Mass

E = K + UE = - GMm/2ro

r

L2 /2mr 2– GMm/r

E < 0

ro = - 2E/(GMm)

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Physics 1710—Chapter 13 Apps: Gravity

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Summary:

The force of attraction between two bodies with mass M and m respectively is proportional to the product of their masses and inversely proportional to the distance between their centers squared.

F = - G M m/ r 2

The proportionality constant in the Universal Law of Gravitation G is equal to 6.673 x 10 –11 N m2 /kg2 .

slide15

Physics 1710—Chapter 13 Apps: Gravity

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Summary:

The gravitational force constant g is equal to

G M/(R+h) 2, R is the radius of the planet.

Kepler’s Laws

The orbits of the planets are ellipses.

The areal velocity of a planet is constant.

The cube of the radius of a planet’s orbit

is proportional to the square of the period.

The gravitation field is the force divided by the mass.

g = Fg / m

slide16

Physics 1710—Chapter 13 Apps: Gravity

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Summary:

The gravitation potential energy for a point mass is proportional to the product of the masses and inversely proportional to the distance between their centers:

U = GMm / r

The escape velocity is the minimum speed a projectile must have at the surface of a planet to escape the gravitational field.

vescape = √[ 2GM/R]

Total Energy E is conserved for two body geavitational problem; bodies are bound for E ≤ 0

E = L2/2mr 2 – GMm/r

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