662. Determine the work required to launch a rocket weighing $P=1.5 T$ from the Earth's surface to a height of $H=2000$ km.
Solution. The force $F$ of attraction of a body by the Earth or the weight of the body depends on its distance $x$ from the center of the Earth: $F(x)=\frac{\lambda}{x^{2}}$, where $\lambda$ is a constant.

If $P$ is the weight of the body when it is on the surface of the Earth, i.e., at a distance of the Earth's radius $R$ from the center of the Earth, then $P=\frac{\lambda}{R^{2}}, \lambda=P R^{2}$ and the force $F$, which the engine of a rising rocket must overcome at the moment when it is at a distance $x$ from the center of the Earth, is a known function of $x$:

$$
F(x)=\frac{P R^{2}}{x^{2}}
$$

Assuming that the work done by the rocket's engine when lifting it to a height $x$ is some function $q(x)$ and allowing that when the rocket is further lifted by a small height $d x$, the force $F$ remains constant, we find the approximate value of the increment of work

$$
\Delta q \approx F(x) d x=\frac{P R^{2}}{x^{2}} d x=d q
$$

When the rocket is lifted from the surface of the Earth to a height $H$, the variable $x$ changes from $R$ to $R+H$. Therefore, the required work $Q$ is expressed by the integral

$$
Q=\int_{R}^{R+H} F(x) d x=P R^{2} \int_{R}^{R+H} \frac{d x}{x^{2}}=\left.P R^{2}\left(-\frac{1}{x}\right)\right|_{R} ^{R+H}=\frac{P R H}{R+H}
$$

For $P=1.5 T, H=2000 \kappa m, R=6400 \kappa . \mu \quad Q \approx 2285714000 \kappa \Gamma m \approx$ $\approx 22422854340 \partial$ N.

The work that the engine must perform to completely free the rocket from Earth's gravity can be determined as the limit of work $Q(H)$ as $H$ increases without bound:

$$
\lim _{H \rightarrow \infty} Q(H)=\lim \frac{P R H}{R+H}=\lim \left[P R:\left(\frac{R}{H}+1\right)\right]=P R
$$

For the given values of $P$ and $R$, this work will be $9600000000 \kappa \Gamma_{\mathcal{M}} \approx 94176000000 \partial$ J.