GravitationClass 11 Physics Notes

Introduction

From a young age, we notice that objects are pulled towards the Earth. A ball thrown up will always come down, and walking uphill is much harder than walking downhill. This pull is due to gravity. The Italian physicist Galileo Galilei (1564-1642) was one of the first to realize that all objects, no matter their mass, accelerate towards the Earth at a constant rate. He famously conducted experiments, like rolling balls down inclined planes, to measure this acceleration due to gravity.

At the same time, people have always been fascinated by the movement of stars and planets. Early models of the universe were often geocentric, meaning they placed the Earth at the center. The most famous of these was proposed by Ptolemy around 2000 years ago, suggesting that the Sun, stars, and planets all revolved around the Earth in complex circular paths.

However, a more accurate model, the heliocentric model, which places the Sun at the center, was suggested much earlier by the Indian astronomer Aryabhatta in the 5th century A.D. A thousand years later, a Polish monk named Nicolas Copernicus (1473-1543) proposed a definitive heliocentric model where planets moved in simple circles around the Sun. This idea was controversial and faced opposition, but it was supported by scientists like Galileo.

The final pieces of the puzzle came from the meticulous work of Danish nobleman Tycho Brahe (1546-1601), who spent his life recording the positions of planets. His assistant, Johannes Kepler (1571-1640), analyzed this vast amount of data and discovered three fundamental laws of planetary motion. These laws, now known as Kepler's laws, were crucial for Isaac Newton to later develop his universal law of gravitation.

Kepler's Laws

Johannes Kepler formulated three laws that describe how planets move around the Sun.

1. Law of Orbits

All planets move in elliptical orbits with the Sun located at one of the two foci of the ellipse.

An ellipse is an oval shape. You can draw one by fixing two pins (the foci, $F_1$ and $F_2$) on a board, looping a string around them, and tracing a path with a pencil while keeping the string taut. A circle is just a special type of ellipse where both foci are at the same point (the center).

• Perihelion: The point in the orbit where the planet is closest to the Sun (Point P in Fig. 7.1a).
• Aphelion: The point in the orbit where the planet is farthest from the Sun (Point A in Fig. 7.1a).
• Semi-major axis: Half the distance between the perihelion and the aphelion.

2. Law of Areas

The line joining a planet to the Sun sweeps out equal areas in equal intervals of time.

This means that a planet moves faster when it is closer to the Sun (at perihelion) and slower when it is farther away (at aphelion). This law is a direct consequence of the conservation of angular momentum. Since gravity is a central force (always pointing towards the Sun), the angular momentum of a planet in its orbit remains constant. The rate at which the area is swept is given by:

$\frac{\Delta A}{\Delta t} = \frac{L}{2m}$

where $L$ is the angular momentum and $m$ is the mass of the planet. Since $L$ and $m$ are constant, the rate of sweeping area ($\Delta A / \Delta t$) is also constant.

3. Law of Periods

The square of the time period of a planet's revolution ($T$) is directly proportional to the cube of the semi-major axis ($a$) of its elliptical orbit.

Mathematically, this is expressed as: $T^2 \propto a^3$

This means that planets farther from the Sun have much longer orbital periods (years) than planets closer to the Sun. The table below shows data for planets in our solar system, confirming this law. The ratio $Q = T^2/a^3$ is nearly constant for all planets.

Given

• Speed at perihelion = $v_P$
• Distance at perihelion = $r_P$
• Speed at aphelion = $v_A$
• Distance at aphelion = $r_A$
• Mass of the planet = $m_p$

To Find

• The relationship between {$r_P, v_P$} and {$r_A, v_A$}.
• Compare the time taken to traverse arc BAC with arc CPB.

Formula

Conservation of angular momentum: $L = mvr$ (when velocity is perpendicular to the radius vector). At perihelion and aphelion, the velocity vector is perpendicular to the position vector from the Sun. $L_P = m_p r_p v_p$ $L_A = m_p r_A v_A$

Solution

From the conservation of angular momentum, the angular momentum at perihelion ($L_P$) must equal the angular momentum at aphelion ($L_A$). $m_p r_p v_p = m_p r_A v_A$ Canceling the mass $m_p$ from both sides, we get: $r_p v_p = r_A v_A$ Rearranging this gives the relationship between the speeds and distances: $\frac{v_p}{v_A} = \frac{r_A}{r_p}$

According to Kepler's second law, the planet sweeps equal areas in equal times. Since the area SBAC is larger than the area SBPC, the planet will take a longer time to traverse the arc BAC than the arc CPB.

Final Answer The relationship is $\frac{v_p}{v_A} = \frac{r_A}{r_p}$. The planet will take longer to traverse BAC than CPB.

Universal Law of Gravitation

Legend says that Isaac Newton was inspired by a falling apple to connect the force that pulls objects to the ground with the force that keeps the Moon in orbit around the Earth. He reasoned that the Moon is constantly accelerating towards the Earth (centripetal acceleration), but this acceleration is much smaller than the acceleration of a falling object on Earth's surface.

The Moon's centripetal acceleration is given by: $a_m = \frac{V^2}{R_m} = \frac{4\pi^2 R_m}{T^2}$ where $R_m$ is the radius of the Moon's orbit and $T$ is its period.

Newton found that this acceleration was about 3600 times smaller than the acceleration due to gravity on Earth, $g$. He also noted that the Moon's orbital radius is about 60 times the Earth's radius, and $60^2 = 3600$. This suggested that the force of gravity decreases with the square of the distance.

This led him to propose the Universal Law of Gravitation: Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Mathematically, the magnitude of the force ($F$) between two point masses $m_1$ and $m_2$ separated by a distance $r$ is: $|\mathbf{F}| = G \frac{m_1 m_2}{r^2}$ Here, G is the universal gravitational constant.

In vector form, the force on mass $m_2$ due to mass $m_1$ is: $\mathbf{F} = -G \frac{m_1 m_2}{r^2} \hat{\mathbf{r}}$ The negative sign indicates that the force is attractive, pulling $m_2$ towards $m_1$. By Newton's third law, the force on $m_1$ due to $m_2$ is equal and opposite.

The Principle of Superposition

If there are multiple masses, the total gravitational force on any one mass is the vector sum of the forces exerted on it by all the other masses. Each force is calculated independently.

For a mass $m_1$ being attracted by masses $m_2, m_3, m_4$, the total force is: $\mathbf{F}_1 = \mathbf{F}_{12} + \mathbf{F}_{13} + \mathbf{F}_{14}$

Gravitational Force of Extended Objects

For large objects (like planets), we must consider the force from every point mass within the object. Using calculus, two important results are found for spherical objects:

1. Outside a Shell: The gravitational force exerted by a uniform spherical shell on a point mass outside it is the same as if the entire mass of the shell were concentrated at its center.
2. Inside a Shell: The gravitational force exerted by a uniform spherical shell on a point mass inside it is zero.

Given

• Three masses at vertices A, B, C are each $m$.
• Mass at centroid G is $2m$.
• Distances AG = BG = CG = 1 m.
• The triangle is equilateral.

To Find

(a) The net force on the mass $2m$ at G. (b) The net force on the mass $2m$ at G if the mass at A is doubled to $2m$.

Formula

Newton's Law of Gravitation: $\mathbf{F} = G \frac{m_1 m_2}{r^2} \hat{\mathbf{r}}$ Principle of Superposition: $\mathbf{F}_R = \mathbf{F}_{GA} + \mathbf{F}_{GB} + \mathbf{F}_{GC}$

Solution

Let's set up a coordinate system with G at the origin and mass A on the positive y-axis. The vertices B and C will be in the third and fourth quadrants, respectively. The angle between GB and the negative x-axis is $30^\circ$, and the angle between GC and the positive x-axis is $30^\circ$.

(a) Calculate the net force with equal masses

The individual forces are:

• Force by mass at A: $\mathbf{F}_{GA} = G \frac{m(2m)}{1^2} \hat{\mathbf{j}} = 2Gm^2 \hat{\mathbf{j}}$
• Force by mass at B: $\mathbf{F}_{GB} = G \frac{m(2m)}{1^2} (-\hat{\mathbf{i}}\cos30^\circ - \hat{\mathbf{j}}\sin30^\circ) = 2Gm^2(-\hat{\mathbf{i}}\cos30^\circ - \hat{\mathbf{j}}\sin30^\circ)$
• Force by mass at C: $\mathbf{F}_{GC} = G \frac{m(2m)}{1^2} (\hat{\mathbf{i}}\cos30^\circ - \hat{\mathbf{j}}\sin30^\circ) = 2Gm^2(\hat{\mathbf{i}}\cos30^\circ - \hat{\mathbf{j}}\sin30^\circ)$

The resultant force is the vector sum: $\mathbf{F}_R = \mathbf{F}_{GA} + \mathbf{F}_{GB} + \mathbf{F}_{GC}$ $\mathbf{F}_R = 2Gm^2\hat{\mathbf{j}} + 2Gm^2(-\hat{\mathbf{i}}\cos30^\circ - \hat{\mathbf{j}}\sin30^\circ) + 2Gm^2(\hat{\mathbf{i}}\cos30^\circ - \hat{\mathbf{j}}\sin30^\circ)$ The x-components cancel out: $-2Gm^2\hat{\mathbf{i}}\cos30^\circ + 2Gm^2\hat{\mathbf{i}}\cos30^\circ = 0$. The y-components are: $2Gm^2\hat{\mathbf{j}} - 2Gm^2\hat{\mathbf{j}}\sin30^\circ - 2Gm^2\hat{\mathbf{j}}\sin30^\circ = 2Gm^2\hat{\mathbf{j}} - 2(2Gm^2\hat{\mathbf{j}} \times \frac{1}{2}) = 2Gm^2\hat{\mathbf{j}} - 2Gm^2\hat{\mathbf{j}} = 0$. So, the total force is zero. This is expected due to the symmetry of the arrangement.

Answer for part (a) = $0$

(b) Calculate the net force when mass at A is doubled

Now, the mass at vertex A is $2m$. The force $\mathbf{F}_{GA}$ changes, but $\mathbf{F}_{GB}$ and $\mathbf{F}_{GC}$ remain the same. The new force from A is: $\mathbf{F'}_{GA} = G \frac{(2m)(2m)}{1^2} \hat{\mathbf{j}} = 4Gm^2 \hat{\mathbf{j}}$ The new resultant force $\mathbf{F'}_R$ is: $\mathbf{F'}_R = \mathbf{F'}_{GA} + \mathbf{F}_{GB} + \mathbf{F}_{GC}$ From part (a), we know that $\mathbf{F}_{GB} + \mathbf{F}_{GC} = -2Gm^2 \hat{\mathbf{j}}$. So, $\mathbf{F'}_R = 4Gm^2 \hat{\mathbf{j}} - 2Gm^2 \hat{\mathbf{j}} = 2Gm^2 \hat{\mathbf{j}}$

Answer for part (b) = $2Gm^2 \hat{\mathbf{j}}$

The Gravitational Constant

The value of the universal gravitational constant, G, was first determined experimentally by English scientist Henry Cavendish in 1798. His apparatus, known as a torsion balance, involved a light rod with two small lead spheres suspended by a thin wire. Two large lead spheres were brought near the small ones.

The gravitational attraction between the large and small spheres created a torque, causing the wire to twist. The wire twisted until the restoring torque of the wire balanced the gravitational torque.

• The gravitational force is $F = G \frac{Mm}{d^2}$.
• The gravitational torque is this force multiplied by the length of the rod, $L$.
• The restoring torque of the wire is $\tau\theta$, where $\tau$ is the restoring couple per unit angle and $\theta$ is the angle of twist.

At equilibrium: $G \frac{Mm}{d^2} L = \tau \theta$ By measuring all other quantities ($M, m, d, L, \theta, \tau$), Cavendish was able to calculate G.

The currently accepted value is: $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$

Acceleration Due to Gravity of the Earth

We can think of the Earth as being made of many concentric spherical shells. For a point outside the Earth, the gravitational force is the same as if the Earth's entire mass ($M_E$) were concentrated at its center.

If an object of mass $m$ is on the surface of the Earth (radius $R_E$), the gravitational force on it is: $F = G \frac{M_E m}{R_E^2}$ According to Newton's second law, this force is also equal to $F = mg$, where g is the acceleration due to gravity.

By equating these two expressions for force, we get: $mg = G \frac{M_E m}{R_E^2}$ $g = \frac{G M_E}{R_E^2}$

This equation allows us to calculate the mass of the Earth, $M_E$, since $g$, $R_E$, and $G$ are all known quantities. This is why it's often said that "Cavendish weighed the Earth."

Acceleration Due to Gravity Below and Above the Surface of Earth

The value of $g$ is not constant; it changes with altitude and depth.

Variation with Altitude

For a point mass $m$ at a height $h$ above the Earth's surface, its distance from the center is $(R_E + h)$. The acceleration due to gravity at this height, $g(h)$, is: $g(h) = \frac{G M_E}{(R_E + h)^2}$ As $h$ increases, the value of $g(h)$ decreases.

For heights much smaller than the Earth's radius ($h \ll R_E$), we can use the binomial approximation: $g(h) = g \left(1 + \frac{h}{R_E}\right)^{-2} \approx g \left(1 - \frac{2h}{R_E}\right)$ This shows that for small heights, $g$ decreases linearly with height.

Variation with Depth

Now consider a point mass $m$ at a depth $d$ below the surface. Its distance from the center is $(R_E - d)$. Based on the shell theorem, the outer spherical shell of thickness $d$ exerts no net gravitational force on the mass inside it. The gravitational force is only due to the smaller, inner sphere of radius $(R_E - d)$.

Assuming the Earth has a uniform density $\rho$, the mass of the inner sphere ($M_s$) is: $M_s = M_E \frac{(R_E-d)^3}{R_E^3}$ The force on the mass $m$ is: $F(d) = \frac{G M_s m}{(R_E-d)^2} = \frac{G m}{(R_E-d)^2} \left( M_E \frac{(R_E-d)^3}{R_E^3} \right) = \frac{G M_E m (R_E-d)}{R_E^3}$ The acceleration due to gravity at depth $d$, $g(d)$, is $F(d)/m$: $g(d) = \frac{G M_E}{R_E^3}(R_E - d) = \left(\frac{G M_E}{R_E^2}\right) \frac{R_E - d}{R_E}$ Since $g = \frac{G M_E}{R_E^2}$, we get: $g(d) = g \left(1 - \frac{d}{R_E}\right)$ This shows that $g$ decreases linearly as we go down from the surface. At the center of the Earth ($d=R_E$), $g$ is zero.

Gravitational Potential Energy

The force of gravity is a conservative force, which means the work done by it in moving an object between two points is independent of the path taken. This allows us to define a gravitational potential energy.

For small heights near the Earth's surface, where $g$ is nearly constant, the potential energy ($W(h)$) at a height $h$ is given by: $W(h) = mgh + W_o$ where $W_o$ is a constant, often taken as zero at the surface ($h=0$).

For arbitrary distances from the Earth, we must use the universal law of gravitation. The work done in lifting a particle of mass $m$ from a distance $r_1$ to $r_2$ from the Earth's center is: $W_{12} = \int_{r_1}^{r_2} \frac{G M_E m}{r^2} dr = -G M_E m \left[ \frac{1}{r_2} - \frac{1}{r_1} \right]$ This work done is equal to the change in potential energy, $W(r_2) - W(r_1)$. This leads to the general expression for gravitational potential energy at a distance $r$ from the center of the Earth: $V(r) = -\frac{G M_E m}{r}$ This formula uses the convention that the potential energy is zero at an infinite distance ($r \rightarrow \infty$). The negative sign indicates that gravity is an attractive force.

For a system of multiple particles, the total potential energy is the sum of the potential energies of all possible pairs of particles.

Gravitational Potential is defined as the potential energy per unit mass at a point. $U(r) = \frac{V(r)}{m} = -\frac{G M_E}{r}$

Given

• Four particles, each of mass $m$.
• They are at the vertices of a square of side $l$.

To Find

• The total potential energy of the system, $W(r)$.
• The gravitational potential at the center of the square, $U(r)$.

Formula

Gravitational Potential Energy between two masses: $V = -\frac{G m_1 m_2}{r}$ Gravitational Potential due to a mass: $U = -\frac{G m}{r}$

Solution

The system has six pairs of particles:

• Four pairs along the sides, with distance $l$.
• Two pairs along the diagonals, with distance $\sqrt{2}l$.

The total potential energy is the sum of the potential energies of all pairs: $W(r) = 4 \times \left(-\frac{Gm^2}{l}\right) + 2 \times \left(-\frac{Gm^2}{\sqrt{2}l}\right)$ $W(r) = -4\frac{Gm^2}{l} - \frac{2Gm^2}{\sqrt{2}l} = -\frac{2Gm^2}{l} \left(2 + \frac{1}{\sqrt{2}}\right)$ $W(r) \approx -5.41 \frac{Gm^2}{l}$

The center of the square is equidistant from all four masses. The distance from the center to any vertex is half the diagonal length: $r = \frac{\sqrt{2}l}{2} = \frac{l}{\sqrt{2}}$. The potential at the center is the sum of the potentials from each of the four masses: $U(r) = 4 \times \left(-\frac{Gm}{r}\right) = 4 \times \left(-\frac{Gm}{l/\sqrt{2}}\right)$ $U(r) = -4\sqrt{2} \frac{Gm}{l}$

Final Answer The potential energy of the system is $W(r) = -5.41 \frac{G m^2}{l}$. The potential at the center is $U(r) = -4\sqrt{2} \frac{Gm}{l}$.

Escape Speed

If we throw an object upwards, it falls back to Earth. If we throw it faster, it goes higher before returning. Is it possible to throw it so fast that it never comes back? Yes, and the minimum speed required to do this is called the escape speed.

We can find this speed using the principle of conservation of energy. An object escapes Earth's gravity if it can reach an infinite distance away. Let's assume the object just barely makes it to infinity, meaning its final speed there is zero ($V_f = 0$).

The total energy of the object must be conserved.

• Initial Energy (at a distance $h+R_E$ from the center, with initial speed $V_i$): $E_{initial} = \frac{1}{2}mV_i^2 - \frac{G M_E m}{h+R_E}$
• Final Energy (at infinity, with final speed $V_f=0$): $E_{final} = \frac{1}{2}mV_f^2 + 0 = 0$

For the object to escape, its initial total energy must be greater than or equal to zero. The minimum speed corresponds to the case where the total energy is exactly zero. $\frac{1}{2}m(V_i^2)_{min} - \frac{G M_E m}{h+R_E} = 0$ $\frac{1}{2}m(V_i^2)_{min} = \frac{G M_E m}{h+R_E}$ If the object is launched from the Earth's surface ($h=0$), the escape speed ($v_e$) is: $(V_i)_{min} = v_e = \sqrt{\frac{2 G M_E}{R_E}}$ Since $g = G M_E / R_E^2$, we can also write this as: $v_e = \sqrt{2gR_E}$ For Earth, the escape speed is approximately $11.2 \text{ km/s}$.

Given

• Mass of first sphere = $M$
• Mass of second sphere = $4M$
• Radius of both spheres = $R$
• Center-to-center separation = $6R$
• Mass of projectile = $m$

To Find

The minimum speed $v$ for the projectile to reach the second sphere.

Formula

Conservation of Mechanical Energy: $E = K.E. + P.E. = \text{constant}$ $E = \frac{1}{2}mv^2 - \frac{G M_1 m}{r_1} - \frac{G M_2 m}{r_2}$

Solution

There is a "neutral point" N between the two spheres where the gravitational pull from both spheres cancels out. To reach the second sphere, the projectile only needs enough energy to reach this neutral point. After that, the pull from the more massive sphere ($4M$) will pull it in.

Let the distance of the neutral point N from the center of the first sphere be $r$. At this point, the forces are equal: $\frac{G M m}{r^2} = \frac{G (4M) m}{(6R - r)^2}$ $(6R - r)^2 = 4r^2$ $6R - r = \pm 2r$ This gives two solutions: $r=2R$ or $r=-6R$. The relevant point is between the spheres, so $r=2R$.

Now, we apply the conservation of energy. The initial energy $E_i$ is at the surface of the first sphere (a distance $R$ from its center and $5R$ from the second sphere's center). The final energy $E_N$ is at the neutral point N, where the speed is momentarily zero for the minimum projection speed.

• Initial Energy: $E_i = \frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{G(4M)m}{5R}$
• Energy at Neutral Point: $E_N = 0 - \frac{GMm}{2R} - \frac{G(4M)m}{(6R-2R)} = -\frac{GMm}{2R} - \frac{4GMm}{4R}$

Setting $E_i = E_N$: $\frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{4GMm}{5R} = -\frac{GMm}{2R} - \frac{GMm}{R}$ Canceling $m$ and simplifying: $\frac{1}{2}v^2 = \frac{GM}{R} + \frac{4GM}{5R} - \frac{GM}{2R} - \frac{GM}{R}$ $\frac{1}{2}v^2 = \frac{GM}{R} \left( \frac{4}{5} - \frac{1}{2} \right) = \frac{GM}{R} \left( \frac{8-5}{10} \right) = \frac{3GM}{10R}$ $v^2 = \frac{6GM}{10R} = \frac{3GM}{5R}$ $v = \sqrt{\frac{3GM}{5R}}$

Final Answer The minimum speed of the projectile is $v = \left(\frac{3GM}{5R}\right)^{1/2}$.

Earth Satellites

Earth satellites are objects that revolve around the Earth. The Moon is Earth's only natural satellite. Since 1957, many artificial satellites have been launched for telecommunications, meteorology, and geophysics.

The motion of a satellite is governed by the same principles as planetary motion. For a satellite of mass $m$ in a circular orbit at a height $h$ above the Earth's surface, the gravitational force provides the necessary centripetal force.

• Centripetal Force Required: $F_c = \frac{mV^2}{R_E+h}$
• Gravitational Force Provided: $F_g = \frac{G M_E m}{(R_E+h)^2}$

Equating these two forces: $\frac{mV^2}{R_E+h} = \frac{G M_E m}{(R_E+h)^2}$ Solving for the orbital speed $V$: $V^2 = \frac{G M_E}{R_E+h}$ $V = \sqrt{\frac{G M_E}{R_E+h}}$ This shows that satellites in higher orbits move slower.

Time Period of a Satellite

The time period $T$ is the time taken to complete one orbit. It is the circumference of the orbit divided by the orbital speed. $T = \frac{2\pi(R_E+h)}{V} = \frac{2\pi(R_E+h)}{\sqrt{\frac{G M_E}{R_E+h}}} = \frac{2\pi(R_E+h)^{3/2}}{\sqrt{G M_E}}$ Squaring both sides gives us Kepler's third law for satellites: $T^2 = \frac{4\pi^2}{G M_E} (R_E+h)^3$ This can be written as $T^2 = k(R_E+h)^3$, where $k$ is a constant.

For a satellite orbiting very close to the Earth's surface ($h \approx 0$), the period $T_0$ is: $T_0 = 2\pi\sqrt{\frac{R_E^3}{G M_E}} = 2\pi\sqrt{\frac{R_E}{g}}$ Using values for $R_E$ and $g$, this time period is approximately 85 minutes.

Given

(i) For Phobos orbiting Mars:

• Period, $T = 7 \text{ h, } 39 \text{ min} = (7 \times 3600) + (39 \times 60) = 27540 \text{ s}$
• Orbital radius, $R = 9.4 \times 10^3 \text{ km} = 9.4 \times 10^6 \text{ m}$
• Gravitational constant, $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$

(ii) For Mars and Earth orbiting the Sun:

• Martian orbital radius / Earth orbital radius, $R_{MS}/R_{ES} = 1.52$
• Earth's year, $T_E = 365$ days

To Find

(i) Mass of Mars, $M_m$. (ii) Length of Martian year, $T_M$.

Formula

(i) Kepler's Third Law for satellites: $T^2 = \frac{4\pi^2}{G M_m} R^3$ (ii) Kepler's Third Law for planets: $\frac{T_M^2}{T_E^2} = \frac{R_{MS}^3}{R_{ES}^3}$

Solution

(i) Calculate the mass of Mars Rearranging the formula for $M_m$: $M_m = \frac{4\pi^2 R^3}{G T^2}$ Substitute the given values: $M_m = \frac{4 \times (3.14)^2 \times (9.4 \times 10^6 \text{ m})^3}{6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2} \times (27540 \text{ s})^2}$ $M_m = \frac{4 \times 9.86 \times 830.584 \times 10^{18}}{6.67 \times 10^{-11} \times 7.5845 \times 10^8}$ $M_m = \frac{32758 \times 10^{18}}{50.59 \times 10^{-3}} = 6.47 \times 10^{23} \text{ kg}$ (Note: The textbook calculation seems to have a typo. Using the given numbers, this is the correct result. The book's answer is $6.48 \times 10^{23} \text{ kg}$)

Answer for part (i) $\approx 6.48 \times 10^{23} \text{ kg}$

(ii) Calculate the length of the Martian year From Kepler's third law: $T_M^2 = T_E^2 \left( \frac{R_{MS}}{R_{ES}} \right)^3$ $T_M = T_E \left( \frac{R_{MS}}{R_{ES}} \right)^{3/2}$ Substitute the given values: $T_M = 365 \text{ days} \times (1.52)^{3/2}$ $T_M = 365 \times 1.8738 = 684.0 \text{ days}$

Answer for part (ii) = 684 days

Energy of an Orbiting Satellite

A satellite in orbit has both kinetic energy (due to its motion) and potential energy (due to its position in Earth's gravitational field).

• Kinetic Energy (K.E.) From the orbital speed equation $V^2 = \frac{G M_E}{R_E+h}$, the kinetic energy is: $K.E. = \frac{1}{2}mV^2 = \frac{G M_E m}{2(R_E+h)}$

• Potential Energy (P.E.) The gravitational potential energy at this distance is: $P.E. = -\frac{G M_E m}{R_E+h}$

• Total Energy (E) The total mechanical energy is the sum of the kinetic and potential energies: $E = K.E. + P.E. = \frac{G M_E m}{2(R_E+h)} - \frac{G M_E m}{R_E+h}$ $E = -\frac{G M_E m}{2(R_E+h)}$

• The total energy is negative, which means it is a bound system and cannot escape Earth's gravity.
• The kinetic energy is positive.
• The potential energy is negative.
• The magnitude of the total energy is equal to the kinetic energy: $E = -K.E.$
• The magnitude of the potential energy is twice the magnitude of the kinetic energy: $|P.E.| = 2 \times K.E.$

Given

• Mass of satellite, $m = 400 \text{ kg}$
• Initial orbital radius, $r_i = 2R_E$
• Final orbital radius, $r_f = 4R_E$
• $g = 9.81 \text{ m s}^{-2}$
• $R_E = 6.37 \times 10^6 \text{ m}$

To Find

• Energy required to transfer the satellite, $\Delta E$.
• Change in kinetic energy, $\Delta K$.
• Change in potential energy, $\Delta V$.

Formula

Total Energy: $E = -\frac{G M_E m}{2r}$ Kinetic Energy: $K = \frac{G M_E m}{2r}$ Potential Energy: $V = -\frac{G M_E m}{r}$ Also, $g = \frac{G M_E}{R_E^2}$, so $G M_E = g R_E^2$.

Solution

First, let's find the initial and final energies.

• Initial Energy ($E_i$ at $r_i = 2R_E$): $E_i = -\frac{G M_E m}{2(2R_E)} = -\frac{G M_E m}{4R_E}$
• Final Energy ($E_f$ at $r_f = 4R_E$): $E_f = -\frac{G M_E m}{2(4R_E)} = -\frac{G M_E m}{8R_E}$

The energy required is the change in total energy: $\Delta E = E_f - E_i = \left(-\frac{G M_E m}{8R_E}\right) - \left(-\frac{G M_E m}{4R_E}\right)$ $\Delta E = \frac{G M_E m}{4R_E} - \frac{G M_E m}{8R_E} = \frac{G M_E m}{8R_E}$ Substitute $G M_E = g R_E^2$: $\Delta E = \frac{(g R_E^2) m}{8R_E} = \frac{g m R_E}{8}$ $\Delta E = \frac{9.81 \times 400 \times 6.37 \times 10^6}{8} = 3.125 \times 10^9 \text{ J}$ (Note: The textbook value is $3.13 \times 10^9 \text{ J}$)

Now, let's find the change in kinetic and potential energies.

• Initial K.E.: $K_i = \frac{G M_E m}{4R_E}$

• Final K.E.: $K_f = \frac{G M_E m}{8R_E}$

• Change in K.E.: $\Delta K = K_f - K_i = \frac{G M_E m}{8R_E} - \frac{G M_E m}{4R_E} = -\frac{G M_E m}{8R_E} = -\Delta E$ $\Delta K = -3.13 \times 10^9 \text{ J}$

• Initial P.E.: $V_i = -\frac{G M_E m}{2R_E}$

• Final P.E.: $V_f = -\frac{G M_E m}{4R_E}$

• Change in P.E.: $\Delta V = V_f - V_i = \left(-\frac{G M_E m}{4R_E}\right) - \left(-\frac{G M_E m}{2R_E}\right) = \frac{G M_E m}{2R_E} - \frac{G M_E m}{4R_E} = \frac{G M_E m}{4R_E}$ $\Delta V = 2 \times \Delta E = 2 \times (3.13 \times 10^9 \text{ J}) = 6.26 \times 10^9 \text{ J}$ (Note: The textbook has a calculation error for $\Delta V$, stating it is negative. The potential energy actually increases (becomes less negative) as the satellite moves to a higher orbit. The textbook value is $-6.25 \times 10^9 \text{ J}$, which is incorrect in sign and magnitude.)

Final Answer

• Energy required, $\Delta E = 3.13 \times 10^9 \text{ J}$.
• Change in kinetic energy, $\Delta K = -3.13 \times 10^9 \text{ J}$ (The satellite slows down).
• Change in potential energy, $\Delta V = 6.26 \times 10^9 \text{ J}$ (The potential energy increases).