Consider this problem from my physics workbook:

A spaceship is launched into a circular orbit close to the earth's surface. What additional velocity must now be imparted to the spaceship to overcome the gravitational pull of the earth?


I tried to use the concept of binding energy of closed systems. Here is how my textbook introduces it:

The total mechanical energy (potential + kinetic) of a closed system is negative. The modulus of this total mechanical energy is the binding energy of the system... It is due to this energy that a particle remains attached within a system. If minimum this much energy is given to a particle in any form, the particle no longer remains attached within the system.

I know that the total mechanical energy of a satellite orbiting close to the earth's surface is $\frac{GMm}{2R}$, where $M$ and $R$ are the mass and radius of the earth respectively. Since minimum this much kinetic energy is to be provided to the satellite, $$\frac{1}{2}mv^2=\frac{GMm}{2R} \rightarrow v = \sqrt{\frac{GM}{R}} = \sqrt{gR}$$

However, according to the key, the correct answer is $(\sqrt{2}-1)\sqrt{gR}$. The solution is brief and I am unable to prove it using energy considerations:

The speed of a satellite in a circular orbit close the earth's surface is $v_o = \sqrt{gR}$ and escape velocity is given by $v_e=\sqrt{2gR}$. Therefore, the additional velocity to escape is $v_e-v_o=(\sqrt{2}-1)\sqrt{gR}$.

Could someone please explain to me why my answer is incorrect, and help me prove why the above the solution is true?

  • $\begingroup$ I figured out why my own answer is incorrect -- I was silly to simply add the extra velocity, the proper way is to add the kinetic energy and then calculate the final velocity, subtract the initial velocity from it. I'd still like help on the second part, though. $\endgroup$ Commented Jan 9, 2021 at 10:05
  • $\begingroup$ Wow. I saw this comment just as I finished typing my answer. I hope it’s of some help. $\endgroup$
    – joseph h
    Commented Jan 9, 2021 at 10:28

1 Answer 1


The orbital velocity (object at $r=2R$) can be calculated if we set

$$\frac{1}{2} mv^2 - \frac{GMm}{2R} = 0$$

so that as you point out the velocity is

$$\tag 1 v= \sqrt{\frac{GM}{R}}$$

and also

$$g = \frac{GM}{R^2}$$


$$GM = gR^2$$


$$gR = \frac{GM}{R}$$

and from equation (1) this means

$$v = \sqrt{gR}$$

Since escape velocity is given by

$$v_e = \sqrt{2Rg}$$

then the additional required velocity is

$$V = \sqrt{2Rg} - \sqrt{Rg} = (\sqrt{2} - 1)\sqrt{Rg}$$


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