Law of Universal Gravitation. Satellites. Weightlessness

[$G$] = m$^3$/(kg$\cdot$s$^2$); $M \approx 6 \times 10^{24}$ kg.

Near the Earth's surface, the force of gravity imparts acceleration $g$ to bodies; $mg = G mM/R^2$, whence $M = gR^2/G$, where $R$ is the radius of the Earth.

$F \approx 2.34 \times 10^{-3}$ N.

$F = \frac{1}{7} G Mm \left[\dfrac{8}{d^2} - \dfrac{1}{(d-R)^2}\right]$.

The sought force is found as the difference between the forces that would act on the small sphere from the entire lead sphere and from the part that was removed from the sphere.

$h \approx 13600$ km.

$g \approx 975$ cm/s$^2$.

The acceleration due to gravity is found from the formulas $g_0 = G \frac{M}{R^2}$ and $g = G \frac{M}{(R+h)^2}$. Hence $\frac{g_0}{g} = (\frac{R+h}{R})^2 = 1 + \frac{2h}{R} + \frac{h^2}{R^2} \approx 1 + \frac{2h}{R},$ since $h/R \ll 1$, and therefore $h^2/R^2 \ll 2h/R$. Thus, $g = g_0 \times (\frac{R}{R+h})^2 \approx \frac{g_0 R}{R+2h}.$

Hint. Let's show that the gravitational force acting from an arbitrary spherical shell on a material point inside it is zero. For this, let's divide the spherical shell into thin spherical layers. A cone with a small solid angle with its vertex at this point cuts out elements from the layer, and it can be shown that $\Delta m_1 / (\Delta m_2) = r_1^2 / r_2^2.$ Therefore, the resultant gravitational forces towards these elements are equal to zero. Dividing the entire layer into such elements, we obtain the initial statement. Consequently, only the mass of that part of the Earth enclosed within the sphere of radius $r$ acts on the point.

$v = \sqrt{gR}$.

If the mass of the body falling through the shaft is $m$, then the acceleration $a$ of the body depends on the distance $r$ to the center of the Earth according to the law $a =\frac{F}{m} = \frac{mgr/R}{m} = g \frac{r}{R}$ where $R$ is the radius of the Earth (see problem 6).

The acceleration is maximum at $r = R$ and becomes zero at $r = 0$. This means that the speed of the body is maximum at $r = 0$. After the body passes the center of the Earth, its acceleration is directed opposite to the velocity, and the speed of the body will decrease. To find the speed of the body at $r = 0$, let's use the law of conservation of energy. Assuming that the potential energy of the body is zero at the center of the Earth, we get $mv^2/2 = \Pi$, where $\Pi$ is the potential energy of the body on the surface of the Earth. It is equal to the work that must be done to very slowly move the body from the center to the surface of the Earth. Let's plot the graph of the dependence of the force that needs to be applied to the body on the distance $r$ (see figure). The work done in moving the body is numerically equal to the area of the shaded triangle (see problems 5 and 17 in Chapter: Work, Power, Energy ), therefore
$\Pi = mgR/2\ \text{ and }\ mv^2/2 = mgR/2.$
Hence
$v = \sqrt{gR}.$

In the direction from east to west with a speed $v = 464$ m/s.

The aircraft must be stationary in the coordinate system associated with the fixed stars. For this, the angular velocity of the aircraft must be equal to the angular velocity of the Earth and directed in the opposite direction. When flying over the equator $v = \omega R_E = 2\pi R_E / T,$ where $R_E$ is the radius of the Earth; $T$ is its period of rotation.

The rotational speed of the Earth is used.

$T = \sqrt{6\pi/(\rho G)} \approx 2$ h 41.6 min.

The weight $P$ of a body is the force with which the body acts on a suspension or support that is stationary relative to the body. According to Newton's third law, the support acts on the body with a force $P'$, equal and opposite to $P$. Thus, two forces are applied to bodies at rest on the planet: the force of gravity $F_g$ and the reaction force of the support $P'$. These forces impart the necessary centripetal acceleration to the body during the rotation of the planet around its axis. At the pole and the equator, these forces are directed along the same line. According to Newton's second law $\frac{mv^2}{R} = F_g - P',$ where $v$ is the speed of the rotational motion of the body together with the Earth around its axis. At the pole $v = 0$ and the centripetal acceleration is absent. The reaction force of the support and, consequently, the weight of the body at the pole are equal to the force of gravity: $P_0 = F_g.$ At the equator $v = \dfrac{2\pi R}{T}$, where $T$ is the period of revolution. The weight of the body at the equator $P_e = F_g - m \frac{4\pi^2 R}{T^2}.$ According to the condition, $P_e = 0.5 P_0 = 0.5 F_g,$ then $\frac{4\pi^2 m R}{T^2} = 0.5 F_g.\tag{1}$ According to the law of universal gravitation $F_g = G \frac{mM}{R^2}.$ Here $m$ is the mass of the body; $M$ is the mass of the planet; $R$ is the radius of the planet, but $M = \frac{4}{3} \pi R^3 \rho,$ where $\rho$ is the density of the planet. Substituting the expressions for $F_g$ and $M$ into formula (1), we get $0.5 G \frac{m \cdot 4/3 \pi R^3 \rho}{R^2} = \frac{4\pi^2 m R}{T^2}.$ After simplification, we find $T = \sqrt{6\pi/(\rho G)}.$

$\rho = 30\pi/(T^2G) \approx 180$ kg/m$^3$.

$t \approx 1$ h 25 min.

See problem 10.

Two forces act on the weighed body: the force of gravity $F_g$, directed towards the center of the Earth, and the reaction force of the support $P'$ (see figure); $P'$ is numerically equal to the weight of the body. The resultant $Q$ of the forces $F_g$ and $P'$ imparts centripetal acceleration $a = \omega^2 r$ to the body. Since $r = R \cos \phi$ ($\phi$ is the geographical latitude), then $a = \omega^2 R \cos \phi\quad\text{and}\quad Q = m\omega^2 R \cos \phi$ where $m$ is the mass of the body. In turn, for the force of gravity $F_g$, we can write $F_g = G mM/R^2 = mg_0,$ where $g_0 = G M/R^2$ ($M$ is the mass of the Earth) is a coefficient with the dimension of acceleration, numerically close to the acceleration of free fall at the pole. According to the cosine theorem from the triangle of forces, we get $P' = \sqrt{F_g^2 + Q^2 - 2F_g Q \cos \phi} = F_g \sqrt{1 + (\frac{Q}{F_g})^2 - \frac{2Q}{F_g} \cos \phi}.$ Considering that $Q/F_g \ll 1$, we can neglect the square of this quantity in the radicand. Using the known approximation $\sqrt{1 - \alpha} \approx 1 - \alpha/2$ for $\alpha \ll 1$, we obtain the final expression: $P' \approx F_g (1 - \frac{Q}{F_g} \cos \phi) = m(g_0 - \omega^2 R \cos^2 \phi)$.

336,000 times.

The equations of motion of the Earth around the Sun and the Moon around the Earth will be: $G \frac{m_E M_S}{R_{ES}^2} = m_E \omega_E^2 R_{ES};$ $G \frac{m_M m_E}{R_{ME}^2} = m_M \omega_M^2 R_{ME},$ or $G \frac{M_S}{R_{ES}^3} = \omega_E^2;$ $G \frac{m_E}{R_{ME}^3} = \omega_M^2.$ Dividing the first equation by the second and considering that $\omega = 2\pi/T$, we find $\frac{M_S}{m_E} = \frac{R_{ES}^3}{R_{ME}^3} \frac{T_M^2}{T_E^2}.$

$m_S \approx 2 \times 10^{30}$ kg.

See problem 14.

No. Because in this case, the force of gravity would have a component not lying in the plane of the orbit.

$v = R\sqrt{\dfrac{g}{R+h}$; $T = 2\pi \dfrac{R+h}{R} \sqrt{\dfrac{R+h}{g}}$.

Assume that the satellite moves in a circular orbit, the center of which is at the center of the Earth. In this case, the satellite moves with centripetal acceleration $a_{\text{cp}} = \frac{v^2}{R+h},$ where $v$ is the speed of the satellite in orbit; $R$ is the average radius of the Earth; $h$ is the average altitude of the satellite. This centripetal acceleration is provided to the satellite by the force of gravity $F = G \frac{mM}{(R+h)^2},$ where $M$ is the mass of the Earth; $m$ is the mass of the satellite; $G$ is the gravitational constant. According to Newton's second law, we have $\frac{m\,v^2}{R+h} = G \frac{mM}{(R+h)^2},$ whence $v^2 = G \frac{M}{R+h}.$ The value $G M$ can be found from the following relation: the weight of the body on the surface of the Earth $mg = G \frac{mM}{R^2},$ whence $G M = gR^2.$ Substituting the value $G M$ into the formula for the speed of the satellite, we get $v = R\sqrt{\frac{g}{R+h}}.$ The period of revolution of the satellite $T = 2\pi/\omega,$ where $\omega = v/(R+h)$ is the angular velocity of the satellite in orbit. Consequently, $T = \frac{2\pi(R+h)}{v} = 2\pi \frac{R+h}{R} \sqrt{\frac{R+h}{g}}.$

$\omega = 2\pi/T \approx 0.001$ rad/s; $v = \sqrt[3]{2\pi g R_E^2 / T} \approx 7.4 \times 10^3$ m/s.

The radius of the orbit $R$ can be found from Newton's second law equation $mv^2/R = G \frac{Mm}{R^2}.$ Expressing the linear velocity $v$ in terms of the angular velocity $\omega$, we get $v = \omega R$. Using the relation $G M = gR_E^2$ (see problem 17), we obtain $R = \sqrt[3]{\frac{gR_E^2}{\omega^2}},$ and further, $v = \omega R = \sqrt[3]{gR_E^2 \omega} = \sqrt[3]{2\pi g R_E^2 / T}.$

$R = \sqrt[3]{gR_E^2/\omega^2} \approx 42000$ km; $v = \sqrt[3]{gR_E^2 \omega} \approx 3.1$ km/h.

See the previous problem. The angular velocity of rotation of the satellite must be equal to the angular velocity $\omega$ of the Earth's rotation. From the Earth, it will seem that the satellite moves forward and backward along the meridian. It will appear stationary if its orbit lies in the plane of the equator.

$v_1 \approx 8$ km/s, the same as Earth's.

The first cosmic velocity is the speed of a satellite moving in a circular orbit around a planet near its surface. Therefore $mv_1^2/R = G mM/R^2,$ whence $v_1 = \sqrt{\frac{G M}{R}}.$ Since $M/R = M_E/R_E$, and $G M_E = gR_E^2$, we get $v_1 = \sqrt{gR_E}.$

$v_1 \approx 4$ km/s, i.e., half that of Earth.

1000 kg.

Neglecting the change in the potential energy of the satellite, we can write $A = mv^2/2$. To move in an orbit with radius $R \approx R_E$, the satellite must have a speed $v = \sqrt{gR_E}$ (see problem 17), then $A = mgR_E/2.$ Hence $m = 2A/(gR_E).$

In the first case 1:1, in the second 2:1.

If the potential energy of a body of mass $m$ in the gravitational field is considered zero infinitely far from the Earth (mass $M$), then at a distance $R$ from the center of the Earth it should be equal to $-G Mm/R$ (this formula can be easily obtained based on the analogy of gravitational and electrostatic fields or by considering the work done when moving the body over a small segment $\Delta R$). To lift the satellite to a height $h$ above the Earth's surface, work must be done equal to the change in the potential energy of the satellite: $A = \Pi_{R_E+h} - \Pi_{R_E} = -G \frac{Mm}{R_E+h} - (-G \frac{Mm}{R_E}) = mg \frac{hR_E}{R_E+h}$ (since $G M/R_E^2 = g$). To launch the satellite at this altitude, it needs to be given kinetic energy $mv^2/2$, i.e., perform work $A_1 = mv^2/2.$ For the satellite to rotate in orbit at height $h$ from the Earth, it needs to be given a speed $v = R_E \sqrt{g/(R_E+h)}$ (see problem 17), then $A_1 = \frac{1}{2} \frac{mgR_E^2}{R_E+h}.$ The ratio of the works is $A/A_1 = 2h/R_E.$ For $h = 3200$ km $A/A_1 = 1$, for $h = 6400$ km $A/A_1 = 2$.

The astronaut will be in a state of weightlessness from the moment the spacecraft engines are turned off in outer space.

Let us consider the forces acting on the spacecraft and on the bodies located within it. The spacecraft and the bodies within it always move with the same acceleration. This acceleration for the spacecraft is determined by the thrust force $\vec{F}_T$, the drag force $\vec{F}_d$, and the resultant gravitational force $\vec{F}_g$, which is proportional to its mass $M$. For any body inside the spacecraft, the acceleration is determined by the normal force $\vec{N}$ (equal in magnitude to the force exerted on the support) and the resultant gravitational force $\vec{F}'_g$, which is proportional to the body's mass $m$: $ \vec{a} = \frac{\vec{F}_T + \vec{F}_d + \vec{F}_g}{M} = \frac{\vec{N} + \vec{F}'_g}{m} $ However, the gravitational force per unit mass (the gravitational field) is the same for both the spacecraft and the body: $ \frac{\vec{F}_g}{M} = \frac{\vec{F}'_g}{m} $ From this, it follows that: $ \vec{N} = \frac{m}{M}(\vec{F}_T + \vec{F}_d) $ Consequently, the force exerted by the body on the support (the "apparent weight" of the body) is: $ \vec{P} = -\vec{N} = -\frac{m}{M}(\vec{F}_T + \vec{F}_d) $ This relationship is valid for any stage of the spacecraft's motion. It can also be used to determine the g-forces experienced by the cosmonaut during takeoff and during entry into the dense layers of the atmosphere. When the spacecraft enters outer space ($\vec{F}_d = 0$) and its engines are shut down ($\vec{F}_T = 0$), then $\vec{P} = 0$. That is, all bodies on the spacecraft are in a state of weightlessness and do not experience the interactions and internal stresses associated with weight. 

Pendulum clocks will stop after the rocket engine is turned off. The movement of the spacecraft will not affect the running of wristwatches.

Many methods are possible, based on the application of Newton's second and third laws. For example:

1. measure the ratio of the velocities acquired by bodies with known and unknown masses after burning the thread compressing a spring between the bodies;
2. act on the body with a known force $F$ and measure the acceleration $a$ obtained by the body in this case, then $m = F/a$;
3. measure the tension force of a pre-calibrated spring when rotating the body attached to this spring with a known constant angular velocity, etc.

Yes, by rotating the spacecraft or turning on the engines.

Yes. The force of gravity acts on bodies in a state of weightlessness (see problem 24). The change in potential energy is still given by the formula $\Delta \Pi = mg_h \Delta h,$ where $g_h$ is the acceleration due to gravity at the given altitude.

Formally, both laws are applicable, but Archimedes' law loses its meaning, since both the body and the displaced liquid are weightless.

The linear and angular velocities of the satellite will increase.

The total energy of the satellite is $E = \Pi + K.$ The potential energy of the satellite $\Pi = -G mM/r$ (see problem 23). We find the kinetic energy $K$ from the relation $\frac{mv^2}{r} = G \frac{mM}{r^2},\quad T = G \frac{mM}{2r}.$ Thus, the total energy $E = -G \frac{mM}{2r}.$ Under conditions of motion with weak friction, the work done against friction forces is produced at the expense of the total energy $E$ of the satellite; in this case, as seen from the last relation, $|E|$ must increase, which is possible if $r$ decreases, and consequently $v$ and $\omega$ (see problem 17) increase.

When falling to Earth. When launched, its speed in the denser layers of the atmosphere near the Earth's surface is low, while during descent it is high.