Why am I getting a negative sign in this derivation?

Question

I want to derive a formula for calculating the final velocity of an object that falls from rest as it reaches the surface of the Earth. I want to take the change in acceleration into account and hence use calculus.

Let $R$ be the radius of the Earth.

Let $h$ be the height above the surface of the Earth.

Assuming $g$ is constant gives:

$$V=\sqrt{2GM\left(\frac{h}{R^2}\right)}$$

But using calculus:

$$a=\frac{\text dv}{\text dt}=\frac{\text dx}{\text dt}\frac{\text dv}{\text dx}=v \frac{\text dv}{\text dx}$$

$$v\cdot \text dv=a\cdot\text dx$$

$$a=\frac{GM}{r^2}$$

$$\int_0^V v\,\text dv=GM \int_{R+h}^R \frac{1}{x^2}\,\text dx$$

$$\frac{V^2}{2}=GM\left(\frac{-h}{R(R+h)}\right)$$

$$V=\sqrt{-2GM\left(\frac{h}{R(R+h)}\right)}$$

Why am I getting that minus sign in the calculus derivation ?

Answer 1

Let the object be the system under consideration and $\hat r$ the unit radial vector point outwards from the centre of the Earth.

The attractive gravitational force on the object due to the Earth $\vec F = \frac{GMm}{R^2} (-\hat r) = -\frac{GMm}{R^2} (\hat r) $ is the external force acting on the system.

$-\frac{GMm}{R^2}$ is the component of the gravitational force in the $\hat r$ direction.

The displacement of this force is $\delta \vec r = \delta r (\hat r)$ where $\delta r$ is the component of the displacement in the $\hat r$ direction.

The work done by the gravitational force undergoing a small displacement $\delta \vec r$ is $\vec F \cdot \delta \vec r = - \frac{GMm}{R^2}\, \delta r$.

The sign of this quantity depends on the sign of the component of the displacement in the $\hat r$ direction.

So the total work done by the gravitational force is $\displaystyle \int^{\rm R}_{\rm R+h} -\frac{GMm}{R^2}\,dr$ which works out to be a positive quantity.
So positive work is done on the object and this is equal to its change in kinetic energy.

Stepping back from this integral you have a negative quantity $-\frac{GMm}{R^2}$ multiplied by a series of negative quantities $\delta r$ which will generate a positive value for the work done on the object by the gravitational force due to the Earth.

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The other thing to note is that you are correct in writing the left hand side integral as $\int_0^V v\,dv$ because you ahve taken $V$ to be the component of the velocity in the $\hat r$ direction.

If you analysed the motion and found a value for $V$ you would find it to be a negative quantity ie the velocity is in the $-\hat r$ direction.

This is difficult to show in this situation but let me illustrate what happens if the gravitational field strength $\vec g = - g\hat r$ is constant.

The equation of motion of the mass is $m \frac {d\vec v}{dt} = m\vec g \Rightarrow \frac {dv}{dt} \hat r = - g\, \hat r \Rightarrow \frac {dv}{dt} = -g $

From this you get $\displaystyle \int _0^V dv = \int _0^t -g \,dt \Rightarrow V = -g\,t$ and you "know" the final velocity will be downwards.

If $t$ is positive then $V$, the component of velocity on the $\hat r$ direction is negative ie downwards.

Answer 2

As pointed out in the comments, the acceleration is in the $-x$ direction, so you need to include a negative sign in the integrand of your integral on the right side of your equation.

As a side-note that won't effect the results, you also technically should change the upper limit of the left integral to $-V$, since the velocity is pointing in the $-x$ direction and increasing in magnitude. Of course you end up squaring the upper limit there anyway, so if you make this mistake you will not catch it in the final answer.

Answer 3

You have a sign error because you have defined positive velocity to be down in the limits of the integral, while positive radial coordinate is up.

Since gravity is a conservative field, you don't need acceleration and forces. You can move straight to energy, which does the integration for you:

$$ T_f +U_f = T_i + U_i$$

Starting at rest ($T_i=0$) becomes: $$ \frac 1 2 mv^2 = -GMm[\frac{1}{R+h}-\frac{1}{R}]$$