I have already answered a & b but I am stuck on c-f. I have provided the correct

Question

I have already answered a & b but I am stuck on c-f. I have provided the correct answers but can someone show me the steps to get there?

(a) What is the speed of a 214 kg satellite in an approximately circular orbit 648 km above the surface of Earth? Answer: 7540 m/s
(b) What is the period? Answer: 97.5 min

Suppose the satellite loses mechanical energy at the average rate of 1.0 10^5 J per orbital revolution. Adopt the reasonable approximation that the trajectory is a "circle of slowly diminishing radius".

(c) Determine the satellite's altitude at the end of its 1400th revolution. Answer: 4.90x10^5 m
(d) Determine its speed at this time. Answer: 7630 m/s
(e) Determine its period at this time. Answer: 94.2 min
(f) What is the magnitude of the average retarding force on the satellite? Answer: 0.00229 N

Explanation / Answer

a) If the orbit is circular then the gravitational force on the satellite plays the role of the centripetal force needed to keep the satellite turning around the earth: Fc = Fg eq.1 The centripetal force Fc is the gravitational force Fg Fc = mV^2/R eq.2 Fc = Centripetal force m = Mass of the satellite V = Speed of the satellite R = Radius of the satellite's orbit measured from the center of the earth R = Re+H where Re = earth's radius and H = satellite's height from the surface of the earth Fg = G.m.M / R^2 eq.3 M = Mass of the earth = 6 x 10^24 kg G = Universal gravitational constant = 6.7 x 10^ -11 N.M^2/kg^2 From eq.1, one can equate the right side of eq.2 with the right side of eq.3: mV^2/R = Gm.M / R^2 eq.4 Or: V^2 = GM / R Or: V = SQRT(GM / R) eq.5 R = 6.37 x 10^6 m + 0.653 x 10^6 m = 7.02 x 10^6m V = SQRT(6.7 x 10 ^ -11 x 6 x 10^24/7.02 x 10^6) V = SQRT( 57.3x10^6) V = 2.39 x 10^4 m/s = 7.57 x 10^3 m/s b)we have: V = w.R Where w = 2 x Pi / T Pi = 3.14 T = Period in seconds w = V/R 2 x 3.14/ T = V / R T = 2 x 3.14 x R / V eq.6 T = 6.28 x 7.02 x 10^6/ 7.57 x 10^3 T = 5.82 x 10^3 s or: T = (5.82 x 10^3/60) min T = 97 min c) The satellite's total mechanical energy is: E = Ec + Eg = Kinetic plus gravitational energy E = mV^2/2 - GMm/R we know from eq.4 that mV^2/2 = GMm/2R so: E = -GMm/2R Delta(E)/revolution = -1.0105 J Delta(E) = -1.0105 x 1600 rev. Delta(E) = -1617 J For the sake of simplicity let us apply differential calculus dE = d/dR(-GMm/2R). dR dE = (GMm/2R^2). dR dR = [-1617x 2 x (7.02 x 10^6)^2]/(6.7x10^ -11x 6x10^24 x 239) dR = -16.59 x 10^ -1 m = -1.659m The new altitude H' = H + dR H' = 653,000 - 1.7m = 652998,3 m d) dE = (GMm/2R^2)dR But from eq. 5 V = SQRT(GM/R) dV = d/dR[SQRT(GM/R)].dR dV = [1/2xSQRT(GM/R)]. (-GM/R^2).dR dV = [1/2xSQRT(5,74 x 10^7)]. (-8.2) . -1.7 dV = 6.6 x 10^ -5 x 8.2 x 1.7 = 9.2 x 10^ -4 m/s The new speed V' is: V' = V + dV = [7.57 x 10^3 + 9.2 x 10^ -4] m/s e) From eq.6 we have T = 2x3.14xR/ V dT = 6.28 dR / V dT = 6.28 x(- 1.7)/(7.57 x 10^3) = -1,41x10^ -3 s The new period T' = (5.82 x 10^3 - 1,41 x 10^ -3)s f) dE/dt = F.V F = (1/V).dE/dt dE/dt = 1.0105 J /rev. = -1.0105J / (5.82 x 10^3s) F = -0.17x 10^ -3 x [1/(7.57x10^3)] = -0.0224 x 10^ -6 N F = -2.24 x 10^ -8 N

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