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Unformatted text preview: madrid (tmm2353) – HW #32 – Antoniewicz – (56445) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 7) 2.0 points A star of mass 7 × 10 30 kg is located at vectorr S = ( 6 × 10 12 , 2 × 10 12 , ) m . A planet of mass 2 × 10 24 kg is initially lo cated at vectorr P i = ( 4 × 10 12 , 4 × 10 12 , ) m and is moving with an initial velocity of vectorv P i = ( 2000 , 14000 , ) m / s . At a time 1 × 10 6 s later, what is the new velocity of the planet? Start by finding the x component, v P f,x . Correct answer: 2041 . 27 m / s. Explanation: In order to use the Momentum Principle to find the new velocity, we have to first find the net gravitational force on the planet from the star. The formula for the gravitational force be tween two objects is vector F gr = − Gm 1 m 2  vectorr  2 ˆ r. We know the masses of the objects, so we just need to find the magnitude of the distance between them and the unit vector pointing from the star to the planet. To find the vector pointing from the star to the planet (let’s call it vectorr P S ), we subtract the position of the star from that of the planet: vectorr P S = vectorr P − vectorr S = ( 4 × 10 12 , 4 × 10 12 , ) m − ( 6 × 10 12 , 2 × 10 12 , ) m = (− 2 × 10 12 , 2 × 10 12 , ) m . We need the length of this vector and the unit vector pointing in the same direction. The length (or magnitude) is given by vextendsingle vextendsingle vextendsingle vectorr P S vextendsingle vextendsingle vextendsingle = radicalBig ( − 2 × 10 12 m) 2 + (2 × 10 12 m) 2 = 2 . 82843 × 10 12 m . And we use the magnitude to help us find the unit vector: ˆ r P S = vectorr P S  vectorr P S  = (− 2 × 10 12 , 2 × 10 12 , ) m 2 . 82843 × 10 12 m = (− . 707107 , . 707107 , ) . Now we have everything we need to calcu late the components of vector F gr . Let’s go ahead and find all three components using vector algebra. vector F gr = − Gm S m P  vectorr P S  ˆ r P S = − ( G )(7 × 10 30 kg)(2 × 10 24 kg) (2 . 82843 × 10 12 m) 2 × (− . 707107 , . 707107 , ) = ( 8 . 2537 × 10 19 , − 8 . 2537 × 10 19 , ) N , where G = 6 . 67 × 10 11 m 3 · kg 1 · s 2 . Now that we know the net gravitational force on the planet due to the star, we can use the Momentum Principle, Δ vectorp = vector F net Δ t = vector F gr Δ t, to find the new velocity, using the velocity update formula we can derive from the Mo mentum Principle: vectorp P f = vectorp P i + vector F gr Δ t m P vectorv P f = m P vectorv P i + vector F gr Δ t vectorv P f = vectorv P i + vector F gr m P Δ t ....
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 Fall '08
 Turner
 Force, Mass, Momentum, General Relativity, Velocity

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