<div id=\"tallPI\" class=\"problemIntroduction\">The gravitational pull of the e

Question

<div id="tallPI" class="problemIntroduction">The gravitational pull of the earth on an object is inversely proportional to the square of the distance of the object from the center of the earth. At the earth's surface this force is equal to the object's normal weight <em><img src="http://session.masteringphysics.com/render?tex=%7B%5Crm+mg%7D" alt="{ m mg}" align="middle" /> </em>, where <img src="http://session.masteringphysics.com/render?tex=g" alt="g" align="middle" />=9.8<img src="http://session.masteringphysics.com/render?tex=%7B%5Crm%7B+m%2Fs%7D%7D%5E2" alt="{ m{ m/s}}^2" align="middle" />, and at large distances, the force is zero.</div>
<div class="problemIntroduction">A) What will be its minimum speed as it strikes the earth's surface?</div>
<div class="problemIntroduction">B) If a 20,<span>000</span>-<img src="http://session.masteringphysics.com/render?tex=%7B%5Crm+kg%7D" alt="{ m kg}" align="middle" /> asteroid falls to earth from a very great distance away, how much kinetic energy will it impart to our planet? You can ignore the effects of the earth's atmosphere.&#160;</div>

Explanation / Answer

a)

Potential energy of asteroid at great distance = 0

Potential energy at the earths surface = -GMm/R


= - 6.67 * 10-11 * 5.97 * 1024 * 2 * 104 / (6.37 * 106)

= -12.5 * 1012 J

As whole energy is converted to Kinetic Energy :

1/2 * m * v2 = 0 - (-12.5 * 1012)

v = 3.53 * 104 m/s = minimum speed.

b)

Kinetic Energy Imparted = 12.5 * 1012 J

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