Physics Rocket Problem

Question

I was working on a problem that asks me to give a definite integral for the energy required to lift a 1-kg payload from earth to the moon via rocket. The background information I am given states that: the distance from the earth to the moon is 362,570 km, earth's surface is 6,371 km from it's center, along with the equations: $$E=F*r$$ and $$F(r)= k/r^2$$

I set up my integral as this: $$\int_{6371000}^{362570000} (\frac{9.8}{r^2})r \,dr$$ and was told my function is wrong, but not why it is wrong. I know the units are off, if I'm trying to end in Joules. What I was thinking is that my integral should be: $$\int_{6371000}^{362570000} (\frac{9.8}{r^2})F \,dr$$ Where F is force. The issue then would be where would acceleration come from if it's not explicitly stated in the problem?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{G}}$ and $\ds{\mrm{M}}$ are the Gravitational Constant and the Earth Mass, respectively.

$\ds{\left\{\begin{array}{rl} \ds{\mrm{R_{\oplus}}\mbox{: Earth Radius} =} & \ds{6371.0088\ \mbox{Km}} \\[2mm] \ds{\mrm{GM \over R_{\oplus}^{2}} = g =} & \ds{9.8\ \mrm{m \over sec^{2}}} \\[2mm] \ds{\mrm{d_{EM}:}\mbox{ Distance Earth-Moon} =} & \ds{384400\ \mbox{Km}} \\[2mm] \ds{\mrm{m} =} & \ds{1\ \mbox{Kg}} \end{array}\right.}$

\begin{align} \Delta\varepsilon & = \int_{\mrm{R_{\oplus}}}^{\mrm{R_{\oplus} + d_{EM}}} {\mrm{GMm} \over r^{2}}\,\dd r = \mrm{GMm \over R_{\oplus}} - \mrm{GMm \over R_{\oplus} + d_{EM}} = \mrm{m{GM \over R_{\oplus}^{2}}\,R_{\oplus}} - \mrm{m\,{GMm \over R_{\oplus}^{2}}\,{R_{\oplus}^{2} \over R_{\oplus} + d_{EM}}} \\[5mm] & = \mrm{mgR_{\oplus}\pars{1 - {R_{\oplus} \over R_{\oplus} + d_{EM}}}} = \mrm{mg\,{1 \over 1 + R_{\oplus}/d_{EM}}\,R_{\oplus}} \\[5mm] & = \mrm{1\ Kg\pars{9.8\ {m \over sec^{2}}}\,{1 \over 1 + 6371.0088/384400}\,6371008.8\ m} \approx \bbx{\ds{6.1418 \times 10^{7}\ \mbox{Joule}}} \end{align}

Answer 2

The units being wrong is a big clue. The force is proportional to $\frac 1{r^2}$ as you say. The force at the surface is $9.8N$. You need to scale your force to the radius at the surface, so $F=9.8 (\frac {6371^2}{r^2})$ The quantity in parentheses is dimensionless, so the force comes out in Newtons as you want. For your integral, you need to decide the units of $r$. If you want the work in joules, those are Newton-meters, so $r$ needs to be in meters. That impacts the constants you have in kilometers.