Given: Mearth = 5.98 times 1024 kg Rearth = 6.37 times 106 ? A satellite of mass

Question

Given: Mearth = 5.98 times 1024 kg Rearth = 6.37 times 106 ? A satellite of mass 762 kg is in a circular orbit at an altitude of 528 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 2 km/s. Let the gravitational potential energy be zero at r = infinity. The universal gravitational constant G = 6.67259 times 10-11 Nm2/kg2. What is the total energy of the satellite in orbit? Answer in units of J. What is the total energy of the satellite just before it hits the ground? Answer in units of J. What is the work done by friction? Answer in units of J.

Explanation / Answer

mass of the earth M = 5.98*10^24 kg radius of earth R = 6.37*10^6 m mass of satellite m = 762 kg altitude h = 528 m speed of satellite when it hits the earth surface v = 2 km / s = 2000 m / s orbital speed of the satellite V = [GM / ( R + h) ] where G = gravitational constant = 6.67 *10^-11 Nm^ 2/ kg^ 2 plug the values we get V = 7913 m / s (a). total energy of the satellite in orbit E = ( 1/ 2) mV^ 2 + [ -GMm / ( R + h) ]                                                              = (2.385* 10^10 )- (4.77*10^-10 )                                                              = - 2.385*10^10 J (b). total energy of the satellite just before hits the ground E ' = ( 1/ 2) mv^ 2 +[-GMm / R ]         E ' = (1.524*10^9 )- ( 4.77*10^10)              = -4.618*10^10 J (c). work done by friction W = E - E '                                              = 2.233*10^10 J

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.