6. Question Details SerCP8 7.P045.ssm. My Notes Ask Your Teache A satellite of m

Question

6. Question Details SerCP8 7.P045.ssm. My Notes Ask Your Teache A satellite of mass 205 kg is launched from a site on Earth's equator into an orbit at 190 km above the surface of Earth. (The mass of the Earth is 5.98 × 1024 kg, and the radius of the Earth is 6.38 × 103 km) (a) Assuming a circular orbit, what is the orbital period of this satellite? 3.95 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. s (b) What is the satellite's speed in its orbit? m/s (c) What is the minimum energy necessary to place the satellite in orbit, assuming no air friction? Need Help? Read It Talk to. Tutor Show My Work (Optional) n Details SerCPR7 PO

Explanation / Answer

Let mass of the Earth = M = 5.98*10^24 kg,

radius of the Earth = R = 6.38*10^3 km

mass of the satellite = m = 205 kg,

satellite’s altitude above the Earth’s surface = r = 190 km

At the Earth's surface mg = ymM/R^2

yM = gR^2

b) Centripetal force = Gravitational force

mv^2/(R+r) = ymM/(R+r)^2

v^2 = gR^2 /(R + r)

v = R sqrt(g/(R+r))

= 6.38*10^6 m *sqrt[9.8 m/s^2/(6.38*10^6 m + 190*10^3 m)]

= 7792 m/s

a) T = 2pi/w

= 2pi(R+r)/v

= 2*pi*(6.38*10^3 + 190)*1000/7792

= 5298 s

c) U = Gravitational potential energy

U = -y*Mm/(R+r)

= -gmR^2/(R+r)

Emin = U(in orbit) - U(on Earth's surface) + Ekin

= -gmR^2/(R+r)-(-gmR^2/R)+0.5mv^2

= gmR^2[1/R-1/(R+r)]+0.5mv^2

= gmRr/(R+r)+0.5mgR^2/(R+r)

= [gmR/(R+r)]*(r+R/2)

= [9.8*205*6.38*10^6/(6.38*10^6 + 190*10^3)] * (190*10^3 + 6.38*10^6/2)

= 6.59 *10^9 J

Answer

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