Q.A 400 kg satellite is in a circular orbit of radius $2R_E$ about the Earth. How much energy is required to transfer it to a circular orbit of radius $4R_E$? What are the changes in the kinetic and potential energies?

The key idea here is that rolling without slipping means the sphere has both translational kinetic energy (due to its centre-of-mass motion) and rotational kinetic energy (due to spinning about its centre). When it slips without rolling — that is, it slides without rotating — the rotational part vanishes, and only the translational part survives. So we need to find how much of the $700~\mathrm{J}$ is translational.

For a solid sphere, the moment of inertia about its centre is $I = \frac{2}{5} M R^2$. When it rolls without slipping, the angular speed $\omega$ and the centre-of-mass speed $v$ are related by $v = \omega R$.

1. Write the total kinetic energy for rolling without slipping. The total KE is the sum of translational and rotational parts:

Substitute $I = \frac{2}{5} M R^2$ and $\omega = v/R$:

2. Combine the terms.

So the total kinetic energy when rolling is $\frac{7}{10} M v^2$, and we are told this equals $700~\mathrm{J}$:

3. Find the translational kinetic energy alone. The translational part is $\frac{1}{2} M v^2$. Using $M v^2 = 1000$:

4. Interpret the slipping case. …

When a body revolves around the Earth in a circular orbit, it is continuously accelerating towards the center of the Earth. This acceleration is called centripetal acceleration, and it requires a centripetal force. In the case of an orbiting body, this necessary centripetal force is provided by the gravitational attraction between the Earth and the body. By equating these two forces, we can derive an expression for the orbital velocity or the time period of revolution.

The key idea is that for a stable circular orbit, the gravitational force acting on the satellite must exactly match the centripetal force required to keep it in that orbit.

1. Identify the forces involved:

   • Gravitational Force ($F_g$): This is the attractive force between the Earth (mass $M$) and the orbiting body (mass $m$). According to Newton's Law of Universal Gravitation, it is given by $F_g = \frac{GMm}{r^2}$, where $G$ is the universal gravitational constant and $r$ is the distance from the center of the Earth to the orbiting body.
   • Centripetal Force ($F_c$): This is the force required to keep the body moving in a circular path. It is directed towards the center of the circle and is given by $F_c = \frac{mv^2}{r}$, where $v$ is the orbital speed. Alternatively, in terms of the time period $T$, the orbital speed is $v = \frac{2\pi r}{T}$, so $F_c = m \left(\frac{2\pi r}{T}\right)^2 \frac{1}{r} = \frac{4\pi^2 m r}{T^2}$.

2. Equate the gravitational force to the centripetal force:

   For a stable orbit, these two forces must be equal:

3. Solve for the time period ($T$): We can cancel the mass of the orbiting body ($m$) from both sides and rearrange the equation to solve for $T^2$:

Taking the square root of both sides gives the time period:

> [!FORMULA]
> The time period of a satellite in a circular orbit of radius $r$ around a body of mass $M$ is $T = 2\pi \sqrt{\frac{r^3}{GM}}$. This is a direct consequence of Kepler's Third Law. …

Concept & Intuition

The acceleration due to gravity decreases as we move away from Earth’s surface because the gravitational force follows an inverse-square law with distance from the Earth’s center. At a height $h$ above the surface, the distance from the center is $R_E + h$. So the gravity at that height is $g_h = \frac{GM}{(R_E + h)^2}$. Comparing two heights means comparing these inverse squares. The given ratio $16:9$ tells us that the gravity at $h_1$ is larger than at $h_2$, so $h_1$ must be smaller than $h_2$. Since $h_1 = 2R_E$ is already large, $h_2$ will be even larger — but we need the exact value.

Step-by-step solution

1. Write the formula for gravity at height $h$ The acceleration due to gravity at a height $h$ above Earth’s surface is:

where $G$ is the gravitational constant and $M$ is Earth’s mass.

At the surface ($h=0$), $g_0 = \frac{GM}{R_E^2}$. So we can also write:

2. Set up the given ratio We are told:

Substitute the expression:

The $g_0$ and $R_E^2$ cancel, leaving:

3. Take square roots Since all distances are positive, we take the positive square root: $$$$

Orbital (translational) kinetic energy. The Moon's orbital speed is $v=\dfrac{2\pi R}{T}$, so

Spin (rotational) kinetic energy. Treating the Moon as a uniform solid sphere, $I=\dfrac{2}{5}Mr^2$, and it spins with $\omega=\dfrac{2\pi}{T}$: …

The key is to connect angular momentum $L$ to the given changes in frequency and kinetic energy. For a particle in uniform circular motion, angular momentum is $L = I\omega$, where $I = mr^2$ is the moment of inertia and $\omega = 2\pi f$ is the angular velocity. Kinetic energy is $K = \frac12 I\omega^2$. By expressing $L$ in terms of $K$ and $\omega$ (or $f$), we can see how it scales when those quantities change.

1. Write the basic relations. For a particle of mass $m$ moving in a circle of radius $r$ with angular speed $\omega$:

2. Express angular momentum in terms of kinetic energy and angular velocity. From $K = \frac12 I\omega^2$, we have $I = \frac{2K}{\omega^2}$. Substituting into $L = I\omega$:

Since $\omega = 2\pi f$, this becomes

So $L$ is directly proportional to $K$ and inversely proportional to $f$.

3. Apply the given changes. Frequency is doubled: $f' = 2f$. Kinetic energy is halved: $K' = \frac12 K$. The new angular momentum is $$$$

Concept — gravitational potential energy of a system. The total PE is the sum over all distinct pairs:

Step 1 — enumerate the pairs. Four equal masses $m$ at the corners of a $3l_0 \times 4l_0$ rectangle give $\binom{4}{2} = 6$ pairs:

• 2 pairs separated by the shorter sides: $r = 3l_0$
• 2 pairs separated by the longer sides: $r = 4l_0$
• 2 pairs along the diagonals: $r = \sqrt{(3l_0)^2+(4l_0)^2} = 5l_0$

Step 2 — sum the pair energies. …

The key idea here is the Stefan–Boltzmann law: the total power radiated by a blackbody (or any body that radiates thermally) is proportional to the fourth power of its absolute temperature. That means if you double the temperature, the power doesn’t just double — it goes up by a factor of $2^4 = 16$. This is a classic result in thermal radiation, and it’s the only physics you need for this problem.

Let’s walk through it step by step.

1. State the Stefan–Boltzmann law For a body at absolute temperature $T$, the radiated power $P$ is given by

where $\sigma$ is the Stefan–Boltzmann constant, $\varepsilon$ is the emissivity (a number between 0 and 1), and $A$ is the surface area. For the same body, $\sigma$, $\varepsilon$, and $A$ are constant — they don’t change when you heat it up.

2. Set up the ratio Since everything except $T$ stays the same, the power is simply proportional to $T^4$:

So for two different temperatures $T_1$ and $T_2$,

3. Plug in the numbers Here $T_1 = 400\,\mathrm{K}$, $T_2 = 800\,\mathrm{K}$, and $P_1 = 1000\,\mathrm{W}$. …