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(a) What is the speed of a 235 kg satellite in an approximately circular orbit 6

ID: 1774425 • Letter: #

Question

(a) What is the speed of a 235 kg satellite in an approximately circular orbit 654 km above the surface of Earth? 7543.18 (b) What is the period? min Suppose the satellite loses mechanical energy at the average rate of 1.4 x 105J 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 1300th revolution. (d) Determine its speed at this time. m/s (e) Determine its period at this time. (f) What is the magnitude of the average retarding force on the satellite? (g) Is angular momentum around Earth's center conserved for the satellite and the satellite Earth system (assuming that system is isolated)? The angular momentum of the satellite is conserved The angular momentum of the satellite Earth system is conserved Both are conserved. Neither are conserved.

Explanation / Answer

a)

Fc = Fg

Fc = mV^2/R

where,

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 = GmM / R^2  

where,

M = Mass of the earth = 6 x 10^24 kg

G = Universal gravitational constant = 6.7 x 10^ -11 N.M^2/kg^2

mV^2/R = Gm.M / R^2

V^2 = GM / R

V = sqrt(GM / R)  

R = 6.37 x 10^6 m + 0.654 x 10^6 m = 7.024 x 10^6m

V = sqrt(6.7 x 10^-11 x 6 x 10^24) / (7.024 x 10^6)

V = 7.56 x 10^3 m/s

b)

T = 2pir / V

T = (2pi x 7.024 x 10^6) / (7.56 x 10^3)

T = 5834.38 s

T = 97.24 min

c)

Use PE = -mMG/r

MG = 3.986 x10^14

For circular orbits, KE = 1/2 PE

E = PE - KE

E = 1/2 PE

E loss through 1300 revolutions

E = -1.4 x 10^5 J x 1300 = - 1.82 x 10^2 J

E = 1/2 mMG(1/(6370km + 654km) - 1/r)

r = 1/(1/(6370km + 654km) - 2E/mMG)

2E/mMG = (2 x 1.82 x 10^2) / (0.235 x 10^6 x 3.986 x 10^14)

2E/mMG = 3.88 x 10^-18

1/(6370km + 654km) = 1.42 x 10^-7

r = 7024 km

altitude = 7024 - 6370 = 654 km

sorry not able to calculate d) and others are depends on it :(