Given: M_earth = 5.98 times 10^24 kg R_earth = 6.37 times 10^6 m A satellite of

Question

Given: M_earth = 5.98 times 10^24 kg R_earth = 6.37 times 10^6 m A satellite of mass 884 kg is in a circular or-bit at an altitude of 517 km above the earth's surface. Because of air friction, the satellite eventually is brought to the earth's surface, it hits the earth with a velocity of 4 km/s. Let the gravitational potential energy be zero at r = infinity. The universal gravitational constant G= 6.67259 times 10^-11 N m^2/kg^2. What is the total energy of the satellite in orbit? Answer in units of J. 006 (part 2 of 3) 10.0 points What is the total energy of the satellite just before it hits the ground? Answer in units of J. 007 (part 3 of 3) 10.0 points What is the work done by friction? Answer in units of J.

Explanation / Answer

Eorbit = -1/2 * [G M m / (Re + A)]

= -1/2 * [6.67 * 10-11 * 5.98 * 1024 * 884 / (6.37 * 106 + 517000)]

= -2.56 * 1010 J

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Eground = - [G M m / Re] + [1/2 m v2]

= - [6.67 * 10-11 * 5.98 * 1024 * 884 / 6.37 * 106] + [1/2 * 884 * 40002]

= -4.82 * 1010 J

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