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# A sky lab of mass m kg is first launched from the surface of the earth in a circular orbit of radius 2R and then it is shefted frm this circular orbit to another circular orbit of radius 3 R. Calculate the energy required :-(a) to place the lab in the first orbit.(b) to shift the lab frm first orbit to the second orbit. (R = 6400 km, g= 10 m/s2)

## 2 Answers

11 years ago

Dear Sanchit,

The energy required to launch a sky lab is the sum of thenergy required to take the lab to that height and the kinetic energy given to it launch in the orbit.

(a)Now the energy required to take the lab to a given height is nothing but the change in the potential energy of the lab

so ΔPE = U(2R) - U(R) = - GMm/2R + GMm/R = GMm/R

now the velocity given to the lab is calculated from the fact that the gravitational force is balanced by centripetal force

mv2/2R = GMm/(2R)2

this gives v= √(GM/2R)

thus the kinetic energy given to it is KE=  m v2/2 = mGM/4R

thus the total energy required to launch into the orbit of 2R radius is ΔPE + KE = 5GMm/4R

(b) Now the total energy of the lab in th orbit of 3R radius is calculated in the similar fashion to get a value of 5GMm/6R

Thus the energy required for this shift is difference of the two total energies = 5GMm/12R

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Adapa Bharath

one year ago
your answer is wrong adapa
part 1 -
u1 + energy = u2 + k2
-GMm/r  + energy =  -GMm/2r  + k2
to find k2,       mv2  =  GMm
2r         (2r)2
k2=  ½ mv2 = GMm                            put in top equation
4r
therefore, energy = GMm/r -GMm/2r + GMm/4r =   3GMm/4r

part 2 -
apply same
u2 + k2 + energy = u3 + k3+
-GMm/2r  + GMm/4r  + energy =  -GMm/3r  + k3
to find k3,       mv2   =  GMm
3r          (3r)2
k3 = ½ mv2 =  GMm/6r                   put in above equation
energy = GMm/4r  – GMm/3r  +  GMm/6r  =  GMm/12r

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