Hmwk_10_Solutions

Hmwk_10_Solutions - Physics 221 Fall 2008 Homework #10...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Physics 221 Fall 2008 Homework #10 Solutions Ch. 12:4-6; Ch. 13 Due Tues, Nov 4, 2008 10.1 A 500 kg satellite is placed in a circular orbit above the earth’s equator such that the satellite always stays directly above the same point on the equator. This is called a geosynchronous orbit. Some communications and other satellites are placed in such orbits in order to provide communications coverage over a large fixed area of the earth’s surface. (a) What is the period of the orbit? By definition, the period of the orbit is T = 24 h = 86400 s . (b) What is the radius of the orbit? How far above the earth’s surface is the satellite, in km and in miles? From Eq. (12.12), we have r = Gm E T 2 4 ± 2 ² ³ ´ µ · 1/3 = (6.67 ¸ 10 ¹ 11 N m 2 /kg 2 )(5.97 ¸ 10 24 kg)(8.64 ¸ 10 4 s) 2 4 2 º » ¼ ½ ¾ ¿ = 4.22 ± 10 7 m = 4.22 ± 10 4 km = 2.62 ± 10 4 mi . Since the radius of the earth is 6.38 ± 10 6 m, the radius of the orbit is 6.61 times the radius of the earth. The height of the satellite above the earth’s surface is 3.58 ± 10 4 km = 2.23 ± 10 4 mi. (c) What is the speed of the satellite? v = 2 r T = 2 (4.22 ² 10 7 m) 8.64 ² 10 4 s = 3.07 km/s . (d) What are the kinetic energy and the potential energy of the satellite? What is the total mechanical energy of the satellite? K = 1 2 mv 2 = 1 2 (500 kg)(3.07 ± 10 3 m/s) 2 = 2.36 ± 10 9 J . U = ± Gm E m r = ± (6.67 ² 10 ± 11 N m 2 /kg 2 )(5.97 ² 10 24 kg)(500 kg) 4.22 ² 10 7 m = ± 4.72 ² 10 9 J E = K + U = 2.36 ± 10 9 J ² 4.72 ± 10 9 J = ² 2.36 ± 10 9 J . We see that the magnitude of the mechanical energy is equal to the kinetic energy, and the magnitude of the potential energy is twice the kinetic energy.
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 (e) At what speed would a small object have to be launched from the satellite in order to escape earth’s gravity, assuming that the satellite is not moving? Compare this escape speed with the speed at which the object would have to be launched from the earth’s surface in order to escape earth’s gravity, again assuming that the earth is not moving. Taking into account the motion of the satellite, does the escape speed from the satellite depend on the direction that the object is launched from the satellite? Explain. Using Example 12.5 on p. 392 in the text, the escape speed of the object from the satellite is v = 2 Gm E r = 2(6.67 ± 10 ² 11 N m 2 /kg 2 )(5.97 ± 10 24 kg) 4.22 ± 10 7 m = 4.34 km/s . The escape speed from the earth’s surface is given in Example 12.5 as 11.2 km/s, about a factor of 2.6 larger than the escape speed from the satellite, as verified by the following: v escape, earth v escape, satellite = r satellite r earth = 42.2 ± 10 3 km 6.38 ± 10 3 km = 2.57 . Since the escape speed is measured with respect to the earth, the speed does depend on which direction the object is launched from the satellite, since the satellite itself is moving at 3.07 km/s. If the object were launched perpendicular to the direction of motion of the satellite, then the escape speed is the same as calculated above. However, if the object is launched in the direction that the satellite is moving, the object would have to be launched only at (4.34 – 3.07) km/s = 1.27 km/s. If the object were launched in the opposite direction to the direction
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 9

Hmwk_10_Solutions - Physics 221 Fall 2008 Homework #10...

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