Homework 7 - Solutions - homework 07 FRENNEA, KYLE Due: Mar...

Info iconThis preview shows pages 1–3. 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: homework 07 FRENNEA, KYLE Due: Mar 7 2007, 4:00 am 1 Question 1 part 1 of 1 10 points On the way from a planet to a moon as- tronauts reach a point where that moons gravitational pull is stronger than that of the planet. The masses of the planet and the moon are, respectively, 5 . 38 10 24 kg and 7 . 36 10 22 kg. The distance from the cen- ter of the planet to the center of the moon is 3 . 29 10 8 m. Determine the distance of this point from the center of the planet. Correct answer: 2 . 94549 10 8 m (tolerance 1 %). Explanation: If r p is the distance from this point to the center of the planet and r m is the distance from this point to the center of the moon, then from the formula GmM p r 2 p = GmM m r 2 m , we obtain q = r m r p = radicalBigg M m M p = radicalBigg 7 . 36 10 22 kg 5 . 38 10 24 kg = 0 . 116963 . On the other hand, r p + r m = R. Eliminating r m from the last two equalities, we obtain r p = R q + 1 = 3 . 29 10 8 m (0 . 116963) + 1 = 2 . 94549 10 8 m . Question 2 part 1 of 1 10 points A satellite moves in a circular orbit around the Earth at a speed of 6 . 7 km / s. Determine the satellites altitude above the surface of the Earth. Assume the Earth is a homogeneous sphere of radius R earth = 6370 km and mass M earth = 5 . 98 10 24 kg . You will need G = 6 . 67259 10 11 N m 2 / kg 2 Correct answer: 2518 . 86 km (tolerance 1 %). Explanation: The gravitational force provides the cen- tripetal acceleration. GmM r 2 = mv 2 r , where M is the mass of the Earth and m is the mass of the satellite. Solving for r yields r = GM v 2 = 8 . 88886 10 6 m Then the height of the satellite above the Earths surface is h = r- R earth = 2518 . 86 km Question 3 part 1 of 2 10 points Given: G = 6 . 67259 10 11 N m 2 / kg 2 . A 1 . 9 kg mass weighs 18 . 62 N on the surface of a planet similar to Earth. The radius of this planet is roughly 5 . 6 10 6 m. Calculate the mass of of this planet. Correct answer: 4 . 60583 10 24 kg (tolerance 1 %). Explanation: By Newtons Law of Universal Gravitation, W = G mM planet r 2 , so M planet = W r 2 Gm = (5 . 6 10 6 m) 2 (6 . 67259 10 11 N m 2 / kg 2 ) 18 . 62 N 1 . 9 kg = 4 . 60583 10 24 kg . homework 07 FRENNEA, KYLE Due: Mar 7 2007, 4:00 am 2 Question 4 part 2 of 2 10 points Calculate the average density of this planet. Correct answer: 6261 . 16 kg / m 3 (tolerance 1 %). Explanation: The volume of the planet is V = 4 3 r 3 so its average density is = M planet V = 3 M planet 4 r 3 = 3 (4 . 60583 10 24 kg) 4 (5 . 6 10 6 m) 3 = 6261 . 16 kg / m 3 . Question 5 part 1 of 1 10 points Given: G = 6 . 673 10 11 N m 2 kg 2 Mars has a mass of about 6 . 40 10 23 kg, and its moon Phobos has a mass of about 9 . 11 10 15 kg....
View Full Document

This homework help was uploaded on 04/18/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 13

Homework 7 - Solutions - homework 07 FRENNEA, KYLE Due: Mar...

This preview shows document pages 1 - 3. Sign up to view the full document.

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