A 287-kg satellite is in circular orbit around the Earth and moving at a speed o

Question

A 287-kg satellite is in circular orbit around the Earth and moving at a speed of 2.41 km/s. How much work must be done to move the satellite into another circular orbit that is twice as high above the surface of the Earth?What is the value of g at the surface of this star?Compare the weight of a 1.80-kg mass on the Earth with its weight on the neutron star. How many times bigger is this mass on the neutron star than on Earth? If a satellite is to circle 12.5 km above the surface of such a neutron star, how many revolutions per minute will it make? Do not enter unit. What is the radius of the geostationary orbit for this neutron star?

Explanation / Answer

we know,

Re = 6.37*10^6 m

Me = 5.98*10^24 kg

let h is the initial height of the sateliite.

radius of Orbit, r = Re + h

we know, orbital speed, vo = sqrt(G*Me/r)

vo^2 = G*Me/r

r = G*Me/vo^2

Re + h = G*Me/vo^2

h = G*Me/vo^2 - Re

= 6.67*10^-11*5.98*10^24/(2410)^2 - 6.37*10^6

= 6.23*10^7 m

r1 = Re + h

= 6.37*10^6 + 6.23*10^7

= 6.867*10^7 m

radius of orbit at twoce height, r2 = Re + 2*h

= 6.37*10^6 + 2*6.23*10^7

= 1.3097*10^8 m

we know, Total energy = -half of potentail energy

so, Workdone = E2 - E1

= -G*Me*m/(2*r2) - (-G*Me*m/(2*r1))

= (G*Me*m/2)*(1/r1 - 1/r2)

= (6.67*10^-11*5.98*10^24*287/2)*(1/(6.867*10^7) - 1/(1.3097*10^8))

= 3.96*10^8 J

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