Is a planet's escape speed the same speed as in a circular orbit around it?

I had this intuitive idea. An object shot upwards with the escape velocity must have the same velocity as the object has when it's in circular orbit at the same height as from where shot upwards starts.

So, if we consider a trajectory straight up, perpendicular to the, say, Earth's surface, with initial speed equal to the escape speed, and change the orbit from perpendicular initial velocity to initial parallel velocity, the trajectory becomes a circle, which essentially is escaping Earth!

Is my intuition right?

Let's test and see!

For a circular orbit, $$r$$ away from the planet's centre, all the centripetal acceleration is supplied by gravity:

$$$$\frac{mv^{2}}{r} = \frac{GMm}{r^{2}} \implies v^{2} = \frac{GM}{r}$$$$

where $$M$$ is the planet mass, $$m$$ is the mass of the orbiting object and $$v$$ is its speed.

On the other hand, if we are to escape the gravitational field of a planet then all the initial kinetic energy we give our body must be used to overcome the gravitational potential energy as we take the object to infinity:

$$$$\frac{1}{2}mv_{e}^{2} = \frac{GMm}{r} \implies v_{e}^{2} = \frac{2GM}{r}$$$$

where $$v_{e}$$ is the escape velocity.

If I've interpreted your question correctly, the answer is an unfortunate no (by a measly factor of $$\sqrt{2}$$, but a factor of $$\sqrt{2}$$ all the same.)

In celestial mechanics, escape velocity or escape speed is the minimum speed needed for a free, non-propelled object to escape from the gravitational influence of a primary body, thus reaching an infinite distance from it.

A circular orbit does not qualify. So escape velocity $$v_e$$ must be greater than the velocity required for a circular orbit $$v_o$$. In fact, you can show that

$$v_e = \sqrt 2 v_o$$

No, but you got close.

The escape velocity is $$v_e=\sqrt{2GM/r}$$ at height $$r$$. You get the factor 2 from the 1/2 in the expression of kinetic energy $$(1/2)mv^2$$ if you equate it to the gravitational potential.

The orbital speed of a circular orbit is $$v_o = \sqrt{GM/r}$$. So $$v_e/v_0 = \sqrt{2}$$: you need to go faster than orbital speed to escape.

This is not strange: imagine that you just instantaneously speed up a bit when in a circular orbit. You will now have an elliptic orbit with a longer semimajor axis, but a perigee where you turned on your engine. To escape to infinity you would need to add some velocity so the apogee diverges to infinity.

• Very clear! Especially the increasing of the speed! The Earth then becomes one of the two foci of an ellipse, which turns to a parabola, if you know what I mean? May 31 at 14:52

Here is a fun fact worth mentioning here:

Imagine you drill a hole all the way through the earth, passing through the center. By physics magic you evacuate the resulting tunnel and, standing near one end of the tunnel, you drop a bowling ball in. It falls freely through the earth, accelerating towards the center and then decelerating all the way up to the other end on the other side of the earth. It then stops, and begins falling back down the tunnel to eventually reappear at your end.

The period of oscillation of the bowling ball is the same as the orbital period of the bowling ball if it were instead fired sideways into an orbit with a radius equal to the earth's radius! We ignore air friction, of course, declare victory, and head out for a beer.

• That would be a hell of a way to transport your ball to the other side! If you arrive in time you can wait for it and grab it out of the air! May 31 at 16:01