The speed of a satellite is v while revolving in an elliptical orbit and is at nearest distance ‘a’ from earth. The speed of satellite at farthest distance ‘b’ will be

To find the speed of the satellite at the farthest distance, we can use the principle of conservation of angular momentum.

A satellite revolving in an elliptical orbit around the Earth is subjected only to a central gravitational force. Since this gravitational force always acts along the line joining the satellite to the center of the Earth, the torque acting on the satellite about the center of the Earth is zero.
Because the net external torque is zero, the angular momentum ($L$) of the satellite remains conserved throughout its orbit.

The magnitude of angular momentum at any point in the orbit is given by:
$L=m·v·r·sin\left(\theta \right)$
where:
• $m$ is the mass of the satellite,
• $v$ is the speed of the satellite,
• $r$ is the distance of the satellite from the center of the Earth,
• $\theta$ is the angle between the velocity vector and the position vector.

At the nearest point (perigee) and the farthest point (apogee) of an elliptical orbit, the velocity vector is perpendicular to the position vector. Therefore, the angle $\theta =90°$, which gives $sin(90°)=1$.

Thus, the angular momentum at these two extreme positions simplifies to:
$L=m·v·r$

Let:
• $v_1=v$ be the speed at the nearest distance,
• $r_1=a$ be the nearest distance,
• $v_2$ be the speed at the farthest distance,
• $r_2=b$ be the farthest distance.

Applying the conservation of angular momentum between the nearest and farthest points:
$m·v_1·r_1=m·v_2·r_2$

We can divide both sides by the mass $m$:
$v_1·r_1=v_2·r_2$

Substitute the given values into the equation:
$v·a=v_2·b$

Solving for the speed at the farthest distance, $v_2$:
$v_2=v\left(\frac{a}{b}\right)$

Therefore, the speed of the satellite at the farthest distance is v (a/b).