Physics, asked by vanijain44081, 10 months ago

Relation between escape velocity and orbital velocity

Answers

Answered by sonuroy76
2

Answer:

Orbital velocity is the speed that an object must maintain in order to orbit another object. Escape velocity is the speed at which an object would break orbit. ... (ll) less than orbital the orbit will decay and the object will crash.

Answered by BrainlyPARCHO
0

  \green{  \fcolorbox{grey}{grey}{ \checkmark \:  \textsf{Verified \: answer}}}

Let v₀ be the orbital (linear) velocity of a body revolving around the Earth, in a circular orbit of radius R.

Kinetic energy = 1/2 m v₀²,

Potential energy in Earth's gravitational field at distance R = - G Me m / R

where, m = mass of the body (or satellite),

Me = mass of Earth, G = Universal Gravitational Constant

The centripetal force for the body in the orbit is supplied by the gravitational force. Hence,

m v₀² / R = G Me m / R² => v₀² = G Me / R ---- equation 1

=> K.E. = 1/2 m v₀² = G Me m / 2 R = - P.E./ 2

Total energy at radius R = KE+PE = - G Me M / 2 R --- equation 2

Suppose now, this body is given an additional velocity v (perpendicular to the orbit and along the radius) such that it goes to a distance d from center of Earth. Since the total mechanical energy is conserved by the gravitational force, the energy at a distance d from the center of Earth is given by:

\begin{gathered}E=\frac{1}{2}mv_d^2-\frac{GM_em}{d}=\frac{1}{2}mv^2-\frac{GM_em}{2R}\\\\For\ d= > \infty,\ and\ v_{\infty}=nearly\ 0,\ minimum\ velocity\ v_e\ needed:\\\\ \frac{1}{2}mv_e^2=\frac{GM_em}{2R}\\\\v_e=\sqrt{\frac{GM_e}{R}}=v_0\\\end{gathered}

The escape velocity of a satellite is the velocity (along radius) required to send it away into the space, just manages to travel to infinite distance. It is equal to the orbital velocity.

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