HW #3-2-solutions - madrid (tmm2353) HW #3-2 Antoniewicz...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: madrid (tmm2353) HW #3-2 Antoniewicz (56445) 1 This print-out should have 16 questions. Multiple-choice 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 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 (lets 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: 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 . Lets go ahead and find all three components using vector algebra. vector F gr = Gm S m P | vectorr P- S | 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, vectorp = vector F net t = vector F gr 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 t m P vectorv P f = m P vectorv P i + vector F gr t vectorv P f = vectorv P i + vector F gr m P t ....
View Full Document

This note was uploaded on 11/15/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

Page1 / 5

HW #3-2-solutions - madrid (tmm2353) HW #3-2 Antoniewicz...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online