Question

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

Answer 1

$\underline{\huge{Answer:-}}$

$\large \bold{Initially,} \\ \: \: \large \mathcal{ E_{i} = - \frac{G \: M_{E} \: m}{8 \: R_{E}} }$

$\large \bold{While \: finally} \\ \: \: \: \: \: \: \: \large \mathcal{E _{f} =- \frac{G \: M_{E} \: m}{8 \: R_{E}} }$

$\large \bold {The \: change \: in \: the \: total \: energy \: is} \\ \large \mathcal { \triangle = E _{f} - E_{i} }$

$\large \mathcal{ \frac{g \: M_{E} \: m}{8 \: R_{E} } = ( \frac{G \: M_{E} }{ R_{E}^{2} }) \frac{m \: R_{E} }{8} }$

$\large \mathcal{ \triangle E = \frac{g \: m \: R_{E} }{8} = \frac{9.18 \times 400 \times 6.37 \times {10}^{6} }{8} = 3.13 \times {10}^{9} \: J }$

The kinetic energy is reduced and it mimics ∆E, namely,

$\large \mathcal{ \triangle K = K_{f} - K_{i} = - 3.13 \times {10}^{9} \: J.}$

The change in potential energy is twice the change in the total energy, namely

$\large \fbox \mathcal{ \triangle V = V_{f} - V_{i} = - 6.25 \times {10}^{9} \: J }$

Answer 2

Answer:

The kinetic energy is reduced and it mimics ∆E, namely,

The kinetic energy is reduced and it mimics ∆E, namely,△K=K

The kinetic energy is reduced and it mimics ∆E, namely,△K=K f

The kinetic energy is reduced and it mimics ∆E, namely,△K=K f

The kinetic energy is reduced and it mimics ∆E, namely,△K=K f −K

The kinetic energy is reduced and it mimics ∆E, namely,△K=K f −K i

The kinetic energy is reduced and it mimics ∆E, namely,△K=K f −K i

The kinetic energy is reduced and it mimics ∆E, namely,△K=K f −K i =−3.13×10

The kinetic energy is reduced and it mimics ∆E, namely,△K=K f −K i =−3.13×10 9

The kinetic energy is reduced and it mimics ∆E, namely,△K=K f −K i =−3.13×10 9 J.

The kinetic energy is reduced and it mimics ∆E, namely,△K=K f −K i =−3.13×10 9 J.

The kinetic energy is reduced and it mimics ∆E, namely,△K=K f −K i =−3.13×10 9 J.

The kinetic energy is reduced and it mimics ∆E, namely,△K=K f −K i =−3.13×10 9 J. The change in potential energy is twice the change in the total energy, namely