Question

EXAMPLE 33. A 400 kg satellite is in a circular
orbit of radius 2 R, about the earth. How much energy
is required to transfer it to a circular orbit of radius
4R? What are the changes in the kinetic and potential
energies?
​

Answer 1

Answer:

Here, m=400,

RE=6.37×106m

Initial total energy, Ei=PE+KE

=−GMmr+12mυ20=−GMmr+12m(GMr)

=−GMm2r=−GMm2×(2RE)=−GMm4RE

Final total energy, Ef=−GMm2×(4RE)=−GMm8RE

The change in total energy is, ΔE=Ef−Ei

−GMm8RE−(−GMm4RE)=GMm8RE

=(GMR2E)×mRE8=gmRE8

=(9.81)×(400)×(6.37×106)8

=3.13×109J Thus, teh energy required to transfer the satllite to the desired orbit =3.13×109J

the decrease in K.E.=Ki−Kf

=12m(GM2RE)−12m(GM4RE)

=18GMmRE=gmRE8=3.13×109J

Gain in P.E.=Uf−Ui

=−GMm4R−(−GMm2R)=GMm4R

=gmRE4=2×3.13×109J=6.26×109J

Thus, the again in P.E. is twice the loss in K.E.

Explanation:

Answer 2

SOLUTION:

Given:

Mass of satellite (m) = 400 kg

The initial energy is given by

$E_{i} = \frac{-GMm}{4R}$

And

Final energy is given by

$E_{f} = \frac{-GMm}{8R}$

Therefore,

The change in the total energy is given by

$\triangle E = E_{f} - E_{i} = \frac{-GMm}{8R} -\frac{-GMm}{4R}$

$\triangle E = \frac{GMm}{8R} = \frac{GM}{R^{2}}(\frac{mR}{8})$

$\triangle E = \frac{gmR}{8} =\frac{9.81* 400*6.37*10^{6}}{8}$

$\boxed{\red{\triangle E= 3.13 \times 10^9J }}}$

The kinetic energy is reduced

Therefore,

$\triangle K = K_{f} - K_{i} = -3.13 \times 10^{3}J$

Change on potential energy,

$\triangle U = -2 \times \tiangle E = -2 \times 3.13 \times 10^{9}$

$\triangle U = -6.25\times 10^{9} J$