Today I wish to relate the physics concepts of energy and gravitation by means of calculus, specifically, integrals. I am often deeply frustrated that IB physics does not offer a calculus based approach to physics. As a result, it misses out on some fundamental connections to mathematics and other areas of physics, as we will see here shortly.
Recently, we learnt in our physics class that $W=Fd$, i.e. work = force $\cdot$ displacement. However, this is only true in the case of constant force. We can still find work as the area under a force * distance graph. To account for variable force then, we can simplify find the area under the curve by integrating the force with respect to the displacement
$$dW=F*dr$$ This equation effectively captures the idea that work done (change in work) is force multiplied by the change in the position (displacement). Here, $r$ is the position vector. Indeed, when we integrate this, we get the general definition of work, $$W=\int_{r_i}^{r_f} F(r) \,dr$$
From Hooke's law, we know that the force due to a spring is given by $F=-kx$ where $k$ is the spring constant and $x$ is the displacement of the spring from its compressed to stretched state.
Substituting this in for the general definition of work above, we see that $$W=\int_{x_i}^{x_f} kx \,dx=\frac{1}{2}kx^2$$ This is the formula we were "told" in class and which we did not get to derive!
For kinetic energy, let's resolve the acceleration and displacement vectors into their components. Here, the key equation to use is $\vec{F}=m\vec{a}$. But first, we will simplify the equation so that it is easier for us to integrate.
Splitting the acceleration and position vectors into their components, $$\vec{F}d\vec{r}=m\vec{a}\ d\vec{r}=m(a_xdx+a_ydy)$$ Using the definition of acceleration as $a=\frac{dv}{dt}$, $$=m(\frac{dv_x}{dt}dx+\frac{dv_y}{dt}dy)$$ Rearranging, $$=m(\frac{dx}{dt}dv_x+\frac{dy}{dt}dv_y)$$ Now using the definition of velocity as $v=\frac{dr}{dt}$, $$=m(v_xdv_x+v_ydv_y)$$ We can now proceed by integrating as follows. $$W=\int_{r_i}^{r_f} F \,dr=\int_{v_i}^{v_f} m(v_xdv_x+v_ydv_y)$$ $$=m[\frac{1}{2}v_x^2+\frac{1}{2}v_y^2]^{v_f}_{v_i}$$ $$=\frac{1}{2}m[(v^2_{x,f}+v^2_{y,f})-(v^2_{x,i}+v^2_{y,i})]$$ but since $\vec{v}=\sqrt{v_x^2+v_y^2},$ $$=\frac{1}{2}m[v_f^2-v_i^2]$$ $$=\frac{1}{2}m\vec{v}$$ Once again, this is the same formula as the one we received in class.
But calculus isn't just useful for deriving physics equations. Recall that an improper integral is unbounded - either one or both endpoints of the definite integral are infinite. To calculate such an improper integral, we can define an arbitrary endpoint (such as $a$, for example) and take the limit of the integral as $a$ approaches $\infty$.
When we have one endpoint approaching positive or negative infinity, $$\int_{x_i}^{\infty} f(x)dx=\lim_{a \rightarrow \infty}{\int_{x_i}^{a} f(x)dx}$$ $$\int_{-\infty}^{x_f} f(x)dx=\lim_{a \rightarrow -\infty}{\int_{a}^{x_f} f(x)dx}$$ When we have both endpoints approaching negative and positive infinity, $$\int_{-\infty}^{\infty} f(x)dx=\lim_{a \rightarrow -\infty}{\int_{a}^{b} f(x)dx}+\lim_{a \rightarrow \infty}{\int_{b}^{a} f(x)dx}$$ where $b$ is just a placeholder variable (recall this important rule for definite integrals) $$\int_{a}^{c} f(x)dx=\int_{a}^{b} f(x)dx+\int_{b}^{c} f(x)dx$$
Finally, we can discuss the key idea of today's post.
According to NASA, the escape velocity is the "speed needed for an object to break away from the gravitational pull of a planet or moon."
As this graphic shows, each planet has a different escape speed.
Recall that Newton's Universal Law of Gravitation states that $F=G\frac{m_1m_2}{r^2}$ where $F$ = force due to gravity, $G$ = gravitational constant, $m_1$ = mass of object 1, $m_2$ = mass of object 2 and $r$ = distance between the centres of the two masses.
The integral below represents the work required to move a mass of $m_1$ from the surface of the Earth ($r_0$ metres from the centre ) to a point completely beyond Earth's gravity. $$W=\int_{r_o}^{\infty} Fdr=\int_{r_o}^{\infty} G\frac{m_1m_2}{r^2}dr$$ Why do we need to set $\infty$ as the upper bound of the definite integral? This is because it is impossible to be totally unaffected by Earth's gravity - no matter how far we go, we will always be influenced by the gravity of Earth.
We can now evaluate this improper integral: $$W=\int_{r_0}^{\infty} Fdr=\int_{r_0}^{\infty} G\frac{m_1m_2}{r^2}dr$$ $$=\lim_{a \rightarrow \infty}Gm_1m_2\int_{r_0}^{a} r^{-2}dr$$ $$=\lim_{a \rightarrow \infty}Gm_1m_2\ \frac{r^{-1}}{-1}|^{a}_{r_0}$$ $$=\lim_{a \rightarrow \infty}Gm_1m_2\cdot [-\frac{1}{a}-(-\frac{1}{r_0})]$$ As $a$ approaches infinity, $\frac{1}{a}$ approaches $0$. $$W=\frac{Gm_1m_2}{r_0}$$ By conservation of energy, the kinetic energy of object 1 must balance the gravitational potential energy we found earlier. Setting them equal, $$\frac{1}{2}m_1v_e^2=\frac{Gm_1m_2}{r_0}$$ Since $m_1$ is cancelled out from the equation altogether, we see that the escape velocity is independent of the object's mass. $$v_e=\sqrt{\frac{2Gm_2}{r_0}}$$ where $G$ is the universal gravitational constant, $m_2$ is the mass of the Earth and $r_0$ is the radius of the Earth. Substituting in values, $$v_e=\sqrt{\frac{2\cdot 6.67 \cdot 10^{-11}m^3kg^{-1}s^{-2}\cdot 5.972\cdot 10^{24}kg}{6.371\cdot10^6m}}$$ $$=11182ms^{-1}=11.18kms^{-1}$$ We see that this value matches the value in the graphic above.
In today's post, we saw how work can be more accurately defined as the integral of force with respect to displacement. We used this definition to derive the formulae for elastic energy and kinetic energy, instead of just being given them. We then saw how we can use the conservation of energy to relate the kinetic energy and the gravitational potential energy of an object exiting the gravity of the Earth, and used improper integrals to calculate the exact escape velocity from the surface of the Earth.
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