Physics 1710 —Warm-up Quiz

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# g for Earth. If the density were 800 kg/m3 - PowerPoint PPT Presentation

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Physics 1710—Warm-up Quiz

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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?

• None of the above

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

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]

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

Physics 1710—Chapter 13 Apps: Gravity

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• Yes, I strongly agree
• Yes, I Agree
• No, I disagree
• No, I strongly disagree

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

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

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

Physics 1710—Chapter 13 Apps: Gravity

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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.

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]??

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]??

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

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)

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 .

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

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