# Ch3 HW1 S2008 - Ch3 HW1 S2008 4/25/08 10:25 AM Web Assign...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Ch3 HW1 S2008 4/25/08 10:25 AM Web Assign Ch3 HW1 S2008 (Homework) Current Score: 21 out of 21 Due: Wednesday, January 30, 200808:35 AM EST Description Gravitational force and prediction of motion. KRISTIN SEILOFF PHYS 2211 M & N Spring 08, section N05, Spring 2008 Instructor: Jennifer Curtis Instructions Reading: Sec. 3.4 - 3.6. In question 5 we lead you through the several steps necessary to calculate the gravitational force as a vector. In question 6 we ask you to do a similar calculation on your own; carry out the same steps as in question 5. 1. 2/2 points The mass of the Sun is 2 10 30 kg, and the mass of Mercury is 3.3 10 23 kg. The distance from the Sun to Mercury is 4.8 10 10 m. (a) Calculate the magnitude of the gravitational force exerted by the Sun on Mercury. 1.919e22 1.92e+22 N (b) Calculate the magnitude of the gravitational force exerted by Mercury on the Sun. 1.919e22 1.92e+22 N Solution or Explanation | grav | = G M 1 M 2 /r2 , therefore the magnitude of the gravitational force is the same for both masses. 2. 5/5 points (a) Calculate the magnitude of the gravitational force exerted by Mercury on a 75 kg human standing on the surface of Mercury. (The mass of Mercury is 3.3 10 23 kg and its radius is 2.4 10 6 m.) 287.891 288 N (b)Calculate the magnitude of the gravitational force exerted by the human on Mercury. 287.891 288 N http://www.webassign.net/v4cgikseiloff3@gatech/student.pl?f=20080425141901kseiloff3@gatech1286165155 Page 1 of 5 Ch3 HW1 S2008 4/25/08 10:25 AM (c) For comparison, calculate the approximate magnitude of the gravitational force of this human on a similar human who is standing 4 meters away. 2.355e - 8 2.36e - 08 N (d) What approximations or simplifying assumptions must you make in these calculations ? ( Note: Some of these choices are false because they are wrong physics!) [x] [x] Treat Mercury as though it were a uniform- density sphere. [_] [_] Ignore the effects of the Sun, which alters the gravitational force that one object exerts on another. [_] [_] Use the same gravitational constant in (a) and (b) despite its dependence on the size of the masses. [x] [x] Treat the humans as though they were points or uniform- density spheres. Solution or Explanation (a) | grav | = G M 1 M 2 /r2 (b) The magnitude of the gravitational force is the same for both masses. (c) | grav | = G M 1 M 2 /r2 (d) The gravitational force that one object exerts on another cannot be altered by other masses. The gravitational constant is not dependent on the size of the masses. 3. 1/1 points A star exerts a gravitational force of magnitude | | on a planet. The distance between the star and the planet is r . If the planet were 3 times farther away (that is, if the distance between the bodies were 3 * r ), by what factor would the force on the planet due to the star change ? For example, if the new force would be 0.01*| | , you should type 0.01 in the box below. Force at distance 3 * r = .111 0.111 *| | Solution or Explanation The gravitational force is proportional to the inverse of the square of the distance r . For example, twice the distance implies 1/4 as big a force; three times the distance, 1/9 as big a force. 4. 1/1 points A planet exerts a gravitational force of magnitude 7 e22 N on a star. If the planet were 5 times closer to the star (that is, if the distance between the star and the planet were 1 / 5 what is is now), what would be the magnitude of the force on the star due to the planet? | | = 1.75e24 1.75e+24 N Solution or Explanation http://www.webassign.net/v4cgikseiloff3@gatech/student.pl?f=20080425141901kseiloff3@gatech1286165155 Page 2 of 5 Ch3 HW1 S2008 4/25/08 10:25 AM The gravitational force is proportional to the inverse of the square of the distance r . For example, half the distance, 4 times as big a force; 1/3 the distance, 9 times as big a force. 5. 4/4 points A planet of mass 4 e24 kg is at location < 2 e11, - 5 e11,0 > m. A star of mass 3 e30 kg is at location < - 5 e11, 2 e11,0> m. It will be useful to draw a diagram of the situation, including the relevant vectors. What is the relative position vector pointing from the planet to the star ? = < - 7e11 - 7.00e+11 , 7e11 7.00e+11 ,0 0 >m What is the distance between the planet and the star ? | | = 9.899e11 9.90e+11 m What is the unit vector = < - .707 - 0.707 , .707 0.707 , -0 0 > What is the magnitude of the force exerted on the planet by the star ? | on planet | = 8.205e20 8.20e+20 N What is the magnitude of the force exerted on the star by the planet? | on star | = 8.205e20 8.20e+20 N What is the force (vector) exerted on the planet by the star ? on planet in the direction of ? = < - 5.801e20 - 5.80e+20 , 5.801e20 http://www.webassign.net/v4cgikseiloff3@gatech/student.pl?f=20080425141901kseiloff3@gatech1286165155 Page 3 of 5 Ch3 HW1 S2008 4/25/08 10:25 AM 5.80e+20 ,0 0 >N What is the force (vector) exerted on the star by the planet? on star = < 5.801e20 5.80e+20 , - 5.801e20 - 5.80e+20 ,0 0 >N Solution or Explanation The relative position vector is "final" minus "initial": calculate the position vector of the star minus the position vector of the planet. The distance between the planet and the star is the magnitude of the relative position vector | |. The unit vector is found by dividing the components of by its magnitude | |. | grav | = GM1 M 2 /r2 , and the magnitude of the gravitational force is the same for both objects. on planet = ( GM1 M 2 /r2 ) ; the vector force exerted on the star by the planet has the same magnitude but the opposite direction (sign). 6. 6/6 points A planet of mass 4 e24 kg is at location < - 7 e11, 2 e11,0 > m. A star of mass 3 e30 kg is at location < 5 e11, - 2 e11,0> m. What is the force exerted on the planet by the star ? (It will probably be helpful to draw a diagram, including the relevant vectors.) = < 4.766e20 4.77e+20 , - 1.589e20 - 1.59e+20 ,0 0 >N Solution or Explanation You first need to find the relative position vector by calculating "final" minus "initial", from the planet toward the star: calculate the position components of the star minus those of the planet. Then divide each component of by the magnitude r = | | to get the unit vector pointing from the planet to the star. Now you can find the vector force: 2 on planet = ( GM 1 M 2 /r ) http://www.webassign.net/v4cgikseiloff3@gatech/student.pl?f=20080425141901kseiloff3@gatech1286165155 Page 4 of 5 Ch3 HW1 S2008 4/25/08 10:25 AM 7. 2/2 points If the mass of a planet is 5.90 surface ? g = 68.628 68.6 N/kg An object of mass 16 kg rests on the surface of this planet. What is the magnitude of the gravitational force on the object ? = 1098.056 1100 N 10 24 kg, and its radius is 2.40 10 6 m, what is the magnitude of the gravitational field, g , on the planet's http://www.webassign.net/v4cgikseiloff3@gatech/student.pl?f=20080425141901kseiloff3@gatech1286165155 Page 5 of 5 ...
View Full Document

## This note was uploaded on 10/19/2009 for the course PHYSICS 2211M taught by Professor Schatz during the Spring '06 term at Georgia Institute of Technology.

Ask a homework question - tutors are online