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Unit 06 Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving
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Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

Jan 05, 2016

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Page 1: Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

Unit 06“Circular Motion,

Gravitation and Black Holes”

Gravitation Problem Solving

Page 2: Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

Gravitational Force Equation

FG = G m1m2

r2

FG = Gravitation Forcem1 = mass of object 1m2 = mass of object 2r = distance between the objects (“r” is measured center to center)G = constant of universal gravitation (6.673x10-11 Nm2/kg2)

Gravitational ForceThe force of attraction between any two masses.

Page 3: Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

Which picture below shows how the radius between two masses should be

measured?

Page 4: Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

r

What happens to the Gravitational Force between two objects if the mass of an object increases?

r

r

If the MASS increases the Gravitational Force also increases!

Page 5: Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

What happens to the Gravitational Force between two objects if the distance between the objects increases?

r

r

If the distance increases the Gravitational Force decreases!!!

The Inverse Square Law says that…. If the radius increased by a factor of 2, the gravitational force decreases by 22 or 4.

Example: above: r1 = 10m F1 = 200N r2 = 20m F2 = 50N

Page 6: Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

The force between two masses is Fg = 9000N. If the radius …

• …increases by a factor of 3. What is the Fg?

• …decreases by a factor of 10. What is the Fg?

The Fg would decrease by 32, or 9. 9000N/9 is 1000N.

Fg = 1000N

The Fg would increase by 102, or 100. 9000Nx100 is 900,000N.

Fg = 900,000N

Page 7: Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

Answer the following questions:1. Are astronauts weightless in outer space?

2. Is there a gravitational force between two people?

3. Is there gravity in outer space?

4. How does the gravitational force change if the mass of an object increases?

5. How does the gravitational force change if the radius of an object decreases?

No, there is gravity. It is just that since the objects are so far away, the gravitational force is very small!

Yes, there is gravity. It is just that since the objects are so small, the gravitational force is very small!

YES! There is gravity. It is just that since the objects are so far away, the gravitational force is very small!

More mass, more gravitational force!

Less radius, more gravitational force!.

Page 8: Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

Easy: What is the gravitational force between two students who are standing 5m away from each other and have masses of 80.0kg and 65.0kg?

Page 9: Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

Easy: What is the gravitational force between two students who are standing 5m away from each other and have masses of 80.0kg and 65.0kg? FG = G m1m2

r2

FG = (6.67x10-11Nm2/kg2) (80kg)(65kg) (5m)2

FG = (6.67x10-11Nm2/kg2) (5200kg2) (25m2)

FG = (6.67x10-11Nm2/kg2) (208kg2/m2)

FG = 1.39x10-8NFG= 0.0000000139N

FG = ?r= 5mm1 = 80kgm2 = 65kg

Page 10: Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

Easy: What is the gravitational force between the sun (m=1.99x1030kg) and the planet Juptier (m=1.89x1027kg) if they are 7.8x1011m apart? FG = G m1m2

r2

FG = (6.67x10-11Nm2/kg2) (1.99x1030kg)(1.89x1027kg)(7.8x1011m)2

FG = (6.67x10-11Nm2/kg2) (3.76x1057kg2)(6.08x1023m2)

FG = 4.12x1023NFG=412,000,000,000,000,000,000,000N

FG = ?r= 7.8x1011mm1 = 1.99x1030kgm2 = 1.89x1027kg

FG = (2.51x1047Nm2)(6.08x1023m2)

Page 11: Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

Easy: What is the escape velocity on the planet Venus which as a mass of 4.90x1024kg and a radius of 6.06x106m?

Page 12: Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

Easy: What is the escape velocity on the planet Venus which as a mass of 4.90x1024kg and a radius of 6.06x106m?

Vesc = 2(6.67x10-11Nm2/kg2)(4.90x1024kg) (6.06x106m)

Vesc = 6.54x1014m3/s2

(6.06x106m)Vesc = 1.08x108m2/s2

Vesc = 10386m/s

Vesc = ?M = 4.90x1024kgR = 6.06x106m

Page 13: Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

What if the Sun were replaced with a Black Hole with the same mass? Would we be pulled into it!?Earth & SunFG = ?mE = 5.97x1024kgmS = 1.99x1030kgr= 1.5x1011m

Earth & Black HoleFG = ?mE = 5.97x1024kgmBH = 1.99x1030kgr= 1.5x1011m

FG = 3.52x1022NFG = 352,000,000,000,000,000,000N

FG = 3.52x1022NFG = 352,000,000,000,000,000,000N

Page 14: Unit 06 “ Circular Motion, Gravitation and Black Holes” Gravitation Problem Solving.

• No, we would not be pulled in! • The gravitational force would be the same because the mass

and distances are the same.• We would orbit around the black hole – just like we orbit

around the sun!!! • We are beyond the event horizon, so we would not get

pulled in. • We would have to be closer, or the black hole would have to

have a larger mass!

What if the Sun were replaced with a Black Hole with the same mass? Would we be pulled into it!?