Question

A 400kg satellite is in a circular orbit of radius 2R about the earth.Calculate the energy needed to transfer it to a circular orbit of radius 4R.Find the changes in it's kinetic energy and it's potential energy (Given R=6400km radius of earth g=10m/s sq.)

Answer 1

Explanation:

Given A 400 kg satellite is in a circular orbit of radius 2R about the earth. Calculate the energy needed to transfer it to a circular orbit of radius 4R. Find the changes in it's kinetic energy and it's potential energy (Given R=6400 km radius of earth g=10m/s sq.)

• Given m = 400 kg, R = 6.4 x 10^6 m
• Initial total energy = Potential energy + kinetic energy
•                            = - Gmm /r + 1/2 mVo^2
•                           = - Gmm / r + 1/2 m((√Gm / r)^2
•                           = - Gmm/2r
•                          = - Gmm / 2(2R)
•                           = - Gmm / 4R
• So final total energy = - Gmm / 2r
•                                = - Gmm / 2(4R)
•                                = - Gmm / 8R
• So change in total energy will be
• ΔE = Ef – Ei
•        = – Gmm / 8R - (- Gmm / 4R)
•        = Gmm / 8R
•      Solving further we get
•   Gm / (R)^2 x m R / 8
•         = gm R / 8
• We know acceleration due to gravity will be g is 10 m / s
•       = 10 x 400 x 6.4 x 10^6 / 8
•     = 3.20 x 10^9 J is the required energy.
• So now decrease in kinetic energy
• K.E = Ki – Kf
•      = 1/2 (Gm / 2R) – 1/2 m(Gm/4R)
• Δ K.E = GmR / 8 = 3.20 x 10^9 J
• Gain in potential energy is Uf – Uc
•                                   = - Gmm / 4R – (- Gmm / 2R)
•                                   = - Gmm / 4R
•                Now solving gmR / 4
•                                 = 10 x 400 x  6.4 x 10 ^6 /  4
•                                 = 6.40 x 10^9 J    

So gain in P.E is twice as loss in K.E

So P.E = - 2K.E

Reference link will be

https://brainly.in/question/4288837

Answer 2

The changes in it's kinetic energy and it's potential energy is

Given:

Mass = m = 400 kg

Radius = R = 6400 km

Acceleration due to gravity = g = 9.8 m/s²

To find:

Kinetic energy and potential energy = ?

Solution:

Initial energy is given by the formula:

$\bold{E_{i}=\frac{-G M_{e}m}{4 R_{e}}}$

Final energy is given by the formula:

$\bold{E_{f}=\frac{-G M_{e}m}{8R_{e}}}$

Change in kinetic energy and potential energy:

$\bold{\Delta E=E_i-E_f}$

$\bold{\Delta E=\frac{-G M_{e} m}{8 R_{e}}+\frac{G M_{e} m}{4 R_{e}}}$

$\bold{\Delta E=\frac{g m R_{e}}{8}}$

On substituting the known values, we get,

$\bold{\Delta E = \frac{9.8\times 400 \times 6400}{8}}$

$\bold{\therefore\Delta E = 3.13 \times 10^9 \ J}$

The Kinetic Energy will become half.

The Potential Energy will become twice.