Question

please give me fast answer

Answer 1

1,3,4 are the answers

Answer 2

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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.