What energy must be given to a spacecraft of mass 1 t in order for it to enter a circular orbit with radius $R = 16000 km$ after starting from the Earth's surface? Consider that the potential energy of a body in the Earth's gravity field at a distance from the Earth's center $R \geq R_{earth}$ is $E_{n} = - \frac{GmM_{earth}}{R}$, where $M_{z}$ is the Earth's mass.
Decision:
In a circular orbit, the total energy of the ship is
$E=\frac{mv^{2}}{2}-G \frac{mM_{earth}}{R}$
Here $v$ is the velocity of the ship in orbit. According to Newton's second law.
$\frac{mv^{2}}{R}=G \frac{mM_{earth}}{R^{2}}=mg \frac{r^{2}_{earth}}{R^{2}}$
(we have taken into account that on the surface of the Earth $mg= G \frac{mM_{earth}}{R^{2}_{earth}}$).) The expression for energy can be rewritten as follows:
$E=- \frac{mgR^{2}_{earth}}{2R}$.
A ship resting on Earth has potential energy
$E_{0}=- G \frac{mM_{earth}}{R_{z}}=-mg R_{earth}$
Thus, to climb into orbit, the ship must be given the energy
$\Delta E = E-E_{0}=mgR_{z}. \left ( 1 - \frac{R_{earth}}{2R} \right ) \approx 5 \cdot 10^{10} J$
Decision:
In a circular orbit, the total energy of the ship is
$E=\frac{mv^{2}}{2}-G \frac{mM_{earth}}{R}$
Here $v$ is the velocity of the ship in orbit. According to Newton's second law.
$\frac{mv^{2}}{R}=G \frac{mM_{earth}}{R^{2}}=mg \frac{r^{2}_{earth}}{R^{2}}$
(we have taken into account that on the surface of the Earth $mg= G \frac{mM_{earth}}{R^{2}_{earth}}$).) The expression for energy can be rewritten as follows:
$E=- \frac{mgR^{2}_{earth}}{2R}$.
A ship resting on Earth has potential energy
$E_{0}=- G \frac{mM_{earth}}{R_{z}}=-mg R_{earth}$
Thus, to climb into orbit, the ship must be given the energy
$\Delta E = E-E_{0}=mgR_{z}. \left ( 1 - \frac{R_{earth}}{2R} \right ) \approx 5 \cdot 10^{10} J$