## Task Condition

Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$.

$$
m=5,0 \mathrm{t}, H=400 \text { km. }
$$
## Solution

By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)

\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction at height } x \\
& F_{0}=G \cdot \frac{m \cdot M}{R_{3}^{2}}=m g-\text { force of attraction at the Earth's surface } \\
& d A=\frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{H} \frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d\left(R_{3}+x\right)=-\left.m \cdot g \cdot R_{3}^{2} \cdot \frac{1}{R_{3}+x}\right|_{0} ^{H}=m \cdot g \cdot R_{3}^{2} \cdot\left(\frac{1}{R_{3}}-\frac{1}{R_{3}+H}\right)= \\
& =5 \cdot 10^{3}[\mathrm{~kg}] \cdot 10\left[\frac{m}{s^{2}}\right] \cdot\left(6380 \cdot 10^{3}[m]\right)^{2} \cdot\left(\frac{1}{6380 \cdot 10^{3}[m]}-\frac{1}{6380 \cdot 10^{3}[m]+400 \cdot 10^{3}[m]}\right)=18820058997 \\
& \text { J] }
\end{aligned}
\]

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Categories: Kuznetsov Problem Book Integrals Problem $22 \mid$ Integrals

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## Problem Kuznetsov Integrals 22-16

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