I had a doubt regarding orbital mechanics. When a satellite burns prograde, ie, in the direction of the instantaneous velocity vector, it increases its orbital speed at the instant, whereas the altitude remains nearly constant(at that instant). However, doesn't this not follow the orbital velocity formula given by $$V = \sqrt{\frac{GM}{R}}$$, which gives a definite speed for a given altitude? This has been confusing me a lot and any explanation will be appreciated!

• – PM 2Ring Jun 25 '20 at 14:48

When its speed suddenly increases its trajectory will not be a circle anymore. Because at distance $$R$$ from center of the planet if satellite's speed is more than $$\sqrt{\frac{GM}{R}}$$ and less than $$\sqrt{\frac{2GM}{R}}$$ its trajectory will be an ellipse. And you can find this ellipse's semi major axis by the formula $$\frac12 mv^2-\frac{GMm}{R}=-\frac{GMm}{2a}$$ and $$v$$ in here is the speed of satellite at distance $$R$$

The formula for orbital speed $$\sqrt{\frac{GM}{R}}$$ is for making satellite's trajectory circle

Hence, the satellite will transfer its orbit to an ellipse

• Specifically, the satellite will transfer to its elliptical orbit at the perigee (lowest point) of the ellipse. – Guy Inchbald Jun 25 '20 at 12:56