Chapter 12: Gravity Selected Solutions

Selected Solutions

7. (a) Earth is pulled in the same direction by the moon and the sun. The two gravitational forces must be added

Therefore, F = 3.56 x 10²² N, toward the sun.

(b) The moon is pulled in opposite directions by the earth and the sun. The sun applies the larger force so we subtract the earth's force from the sun's.

Therefore, F = 2.40 x 10²⁰ N, toward the sun.

(c) Both the earth and the moon apply forces in the same direction on the sun. The two forces must, therefore, be added.

Therefore, F = 3.58 x 10²² N, toward the earth-moon system.

17. The expression for the gravitational acceleration at the surface of a spherical body of mass M and radius R is g = GM/R². So, for Titan,

27. (a) We can use the expression for the period of a satellite in circular orbit (around Earth) of radius r

(b) Since the orbit is circular and moves with a constant speed, the orbital speed is given by the circumference divided by the period. Therefore,

37. The minimum kinetic energy needed to escape is that which gives . Therefore, by energy conservation (taking ) we have

(a) For the moon:

(b) For the earth:

45. Taking h = 110 km, we can obtain the impact speed v_f by using energy conservation.

Selected Solutions by David Reid, Eastern Michigan University. ©2002 by Prentice Hall, Inc.

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