A mistake in the derivation of escape velocity

[Leo Liu]

Homework Statement

NA

Relevant Equations

NA

In the last step of the derivation of escape velocity, the two sides of the equation seem to have opposite signs.
$$-1/2mv_0^2=-mgR_e^2\,\lim_{r\to\infty}(1/r-1/R_e)$$
$$-1/2mv_0^2=mgR_e^2 \frac{1}{R_e}$$
Since the mass and the square of the velocity are positive, the left side of the equation is negative; whereas, the right side of the equation is positive in that g, the radius of the earth, and the mass are all positive.
Please note that g is positive because the author seems to have used a substitution $g=GM_e/R_e^2$.

I would like to know what the cause of the error is and how to avoid it. Thanks!

 

[mpresic3]

I propose ? The integral of the gravitational force over dr calculated (maybe ?) is the work done by the gravitational field in moving the body from R to r. However what may be needed is the work done by the body (not the work done by the field) in overcoming the gravity. ?

 

[Master1022]

I would like to know what the cause of the error is and how to avoid it. Thanks!

Hi,

I do agree with the post above about the definitions used potentially leading to discrepancies. However, I believe they moved the -ve sign from the $1/2 m v_0 ^ 2$ to the integral as the limits have swapped. If you evaluate that integral $1/r^2$ the bracket would be $(1/R_e - 1/r)$. Therefore, I think the author has multiplied through by -1 to swap the order.

In terms of avoiding the error in future, I often get confused with these -ve signs and one way to deal with it is to write out the energy transfer and it will become clearer. For example in this case we have that:
$$ KE_i = Work Done + KE_f $$
where work done is against gravity by the rocket. As it says in the book, we are looking for the lowest amount of kinetic energy that will get us to escape (so we let $KE_f = 0$).
Then we have:
$$ \frac{1}{2} m v_0 ^ 2 = \int_{R_e}^\infty \frac{GMm}{r^2} dr = \int_{R_e}^\infty \frac{gm R_e^2}{r^2} dr $$
The work done by the rocket integral yields a positive value as expected. Then the expression can be re-arranged as required to solve for $v_0$.

Hope that is of some help

 

[TSny]

As @Master1022 pointed to, the mistake is in the last equality of the following

The integral $\large \int_ {_{R_e}}^{^r} \frac{dr}{r^2}$ evaluates to $- \large \left( \frac{1}{r} - \frac{1}{R_e} \right)$ . So the final expression should be $+mgR_e^2 \large \left( \frac{1}{r} - \frac{1}{R_e} \right)$.

 

[mpresic3]

I think the best way to circumvent the error is to use potential energy instead of the work done.

Total Energy at r = R = 1/2 m v0 squared + potential energy = 1/2 m v0 squared + ( - GMm / r ).
Total Energy at r = inf = 1/2 m v squared + potential energy = 1/2 m v squared + ( - GMm / inf) ).

at infinity for escape velocity v (r=inf) =0 so total energy (at r = inf) = 0.

Energy is conserved so Total energy at r = R also = 0

Therefore 1/2 m v0 squared + ( - GMm / r ) = 0; so 1/2 m v0 squared = GMm / r .

 

[Ibix]

It is said that all theoretical physicists make sign errors. Good theoretical physicists make even numbers of sign errors.