362-5 - MAT 362 Answers to selected exercises, week of...

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

View Full Document Right Arrow Icon
MAT 362 Answers to selected exercises, week of Sept. 28 Graded problems 1. The work done is the di f erence in the potential function: W = f ( r 2 ) - f ( r 1 ) , where f ( r ) = - GMm / r . We have G = 6 . 7 × 10 - 8 cm 3 / s 2 g, m = 6 × 10 27 g, M = 3 . 3 × 10 5 m , r 1 = 1 . 5 × 10 12 cm, and r 2 = r 1 + 1. Since r 2 = r 1 + 1 is only slightly larger than r 1 , Taylor’s theorem implies that, to a good approximation, W = - GMm 1 r 2 - 1 r 1 ! GMm 1 r 1 ! 2 . Substituting in the values for the constants yields W 3 . 5 × 10 31 dynes = 3 . 5 × 10 22 J . 2. F is the curl of another vector field only if ∇ · F = 0. (a) ∇ · F = 3, so F is not the curl of another vector field. (b) ∇· F = 2 x - 2 x + 0 = 0, so F is the curl of another vector field. To find it, use the recipe in Exercise 4. This gives G 1 = Z z 0 ( t - 2 xy ) dt - Z y 0 0 dt = z 2 / 2 - 2 xyz G 2 = - Z z 0 ( x 2 + 1) dt = - z ( x 2 + 1) G 3 = 0 . It is straightforward to check that
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

362-5 - MAT 362 Answers to selected exercises, week of...

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