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?

Concept & Intuition

When a body is projected to a height comparable to Earth’s radius, the gravitational force is no longer constant. Using $g$ as the surface value, we must account for the decrease in gravity with distance. The simplest tool is conservation of mechanical energy: the sum of kinetic and gravitational potential energy at launch equals the sum at the maximum height. The potential energy for a radial distance $r$ from Earth’s center is $-\frac{GMm}{r}$, not $mgh$.

1. Set up the energy conservation equation At the surface ($r = R$), the body has kinetic energy $\frac{1}{2} m v^2$ and potential energy $-\frac{GMm}{R}$. At the highest point ($r = R + 10R = 11R$), the kinetic energy is zero, and potential energy is $-\frac{GMm}{11R}$. Energy conservation gives:

2. Simplify the potential energy terms Bring the potential terms together:

3. Replace $GM$ with $gR^2$ At Earth’s surface, $g = \frac{GM}{R^2}$, so $GM = gR^2$. Substitute: $$$$

The key concept is that for a sphere of uniform density, escape velocity is proportional to the radius. This follows from the formula $v_e = \sqrt{\frac{2GM}{R}}$ and the fact that mass $M = \frac{4}{3}\pi R^3 \rho$. Substituting gives $v_e = \sqrt{\frac{8\pi G \rho}{3}} \, R$, so for a fixed density, escape velocity is directly proportional to the radius. Thus, comparing escape velocities reduces to comparing the planets’ radii.

1. Find the radius of planet P.

   Let its radius be $R_P$. Its surface area is $A = 4\pi R_P^2$.

   Its mass is $M_P = \frac{4}{3}\pi R_P^3 \rho$.

2. Find the radius of planet Q.

   Its surface area is given as $4A$. So $4A = 4\pi R_Q^2$.

   But $A = 4\pi R_P^2$, so $4(4\pi R_P^2) = 4\pi R_Q^2$ → $16\pi R_P^2 = 4\pi R_Q^2$ → $R_Q^2 = 4R_P^2$ → $R_Q = 2R_P$.

   Its mass is $M_Q = \frac{4}{3}\pi (2R_P)^3 \rho = 8 \cdot \frac{4}{3}\pi R_P^3 \rho = 8M_P$.

3. Find the radius of planet R.

   Its mass is $M_R = M_P + M_Q = M_P + 8M_P = 9M_P$.

   Since density is the same, $M_R = \frac{4}{3}\pi R_R^3 \rho$ and $M_P = \frac{4}{3}\pi R_P^3 \rho$.

   Dividing: $\frac{M_R}{M_P} = \frac{R_R^3}{R_P^3}$ → $9 = \frac{R_R^3}{R_P^3}$ → $R_R = \sqrt[3]{9} \, R_P$. …

Concept & Intuition

Escape velocity is the minimum speed needed for an object to break free from a celestial body’s gravitational pull. The formula is

where $M$ is mass and $R$ is radius. Notice it scales as $\sqrt{M/R}$. So if we know how mass and radius compare between two bodies, we can directly find the ratio of their escape velocities — no need for actual values.

Step-by-step reasoning

1. Set up the ratio

   Let the moon’s mass be $M_m$ and its radius $R_m$. The planet’s mass is $M_p = 121 M_m$. The planet’s diameter is 9 times the moon’s, so its radius is also 9 times: $R_p = 9 R_m$.

2. Write escape velocities

   For the planet:

For the moon:

3. Find the ratio Divide the planet’s escape velocity by the moon’s: $\frac{v}{v_m} = \sqrt{\frac{121 M_m / (9 R_m)}{M_m / R_m}} = \sqrt{\frac{121}{9}} = \frac{11}{3}$ …

The key idea is that for a satellite in a circular orbit, the total mechanical energy (kinetic + potential) is half the gravitational potential energy and negative. To move to a higher orbit, you must add energy — exactly the difference between the two total energies. This is not simply the change in potential energy, because the kinetic energy also changes (it decreases as you go higher). The net effect is a smaller positive number than you might guess.

Let’s work it out step by step.

1. Total energy in a circular orbit For a body of mass $m$ in a circular orbit of radius $r$ around Earth (mass $M$), the gravitational force provides the centripetal force:

So kinetic energy is:

Gravitational potential energy (taking zero at infinity) is:

Hence total mechanical energy:

This is a standard result: for any circular orbit, $E = -K = \frac12 U$.

2. Energy at the lower orbit (radius $3R$)

3. Energy at the higher orbit (radius $4R$)

4. Energy required to move from lower to higher orbit This is the change in total energy:

Find a common denominator (24):

The key idea is the relationship between orbital kinetic energy and escape energy. In a circular orbit, the gravitational potential energy is $-2K$ (where $K$ is the kinetic energy), so the total mechanical energy is $-K$. To escape, the total energy must become zero (or positive). Therefore, we need to add exactly $K$ more energy — the same amount the satellite already has.

Let’s work through it step by step.

1. Recall the energy of a satellite in a circular orbit. For a satellite of mass $m$ orbiting Earth (mass $M$) at radius $r$, the gravitational force provides the centripetal force:

Hence the kinetic energy is

2. Find the gravitational potential energy. The potential energy (taking zero at infinity) is

Comparing with $K$, we see $U = -2K$.

3. Total mechanical energy in orbit.

So the satellite is bound with total energy $-K$.

4. Condition for escape. To just escape to infinity (with zero speed at infinity), the total energy must be zero:

The satellite already has $E_{\text{orbit}} = -K$. The additional energy $\Delta E$ needed satisfies

The key concept here is escape speed — the minimum speed a rocket needs to break free from a planet’s gravitational pull without further propulsion. It comes from equating kinetic energy to gravitational potential energy:

which gives

Notice that escape speed depends only on the planet’s mass and radius, not on the rocket’s mass. So the ratio for two planets is simply

Now let’s work through it step by step.

1. Write the given data in ratio form.

   Earth’s mass is nine times planet X’s mass: $M_E = 9 M_X$.

   Earth’s radius is four times planet X’s radius: $R_E = 4 R_X$.

2. Set up the escape speed ratio.

3. Substitute the given ratios.

4. Simplify the square root. $\sqrt{\frac{4}{9}} = \frac{2}{3}$ …

Escape velocity: $v_e=\sqrt{\dfrac{2GM}{R}}$, so

Therefore …

Concept: energy conservation. A body projected with speed u > v_e retains kinetic energy far from the Earth.

(1/2)m u^2 - (1/2)m v_e^2 = (1/2) m v_inf^2

v_inf = sqrt(u^2 - v_e^2)

With u = 3 v_e: …

Concept: energetics of a satellite in a circular orbit of radius r.

Gravitational force provides the centripetal force:

GMm/r^2 = mv^2/r => mv^2 = GMm/r

KE = (1/2)mv^2 = GMm/(2r)

PE = -GMm/r …

The key idea here is the relationship between kinetic energy, potential energy, and total mechanical energy for a satellite in a circular orbit. When a satellite orbits close to Earth, its orbit is nearly circular, and we can use the standard orbital energy equations.

For a satellite of mass $m$ in a circular orbit of radius $r$ (where $r$ is approximately Earth's radius $R$ for a low orbit), the gravitational force provides the centripetal acceleration:

From this, the orbital speed is $v = \sqrt{\frac{GM}{r}}$, and the kinetic energy is:

The gravitational potential energy of the satellite is:

So the total mechanical energy is:

This is a crucial result: for a circular orbit, the total energy is negative and exactly equal to $-K$.

Now, to "just overcome" Earth's gravity means the satellite should reach escape speed — the speed at which it can go to infinity with zero kinetic energy remaining. At escape speed, the total mechanical energy becomes zero (kinetic energy exactly balances the negative potential energy at that point). So the satellite needs to go from total energy $-K$ to total energy $0$.

Let's work through the steps:

1. Current state: The satellite has kinetic energy $K$ and total energy $E = -K$. It is bound to Earth.

2. Target state for escape: The satellite needs total energy $E' = 0$. This means its new kinetic energy $K'$ must satisfy $K' + U = 0$, so $K' = -U = \frac{GMm}{r} = 2K$. …

Step 1 — Speed gained in the accelerating field.

All the electrical work done on the charge becomes kinetic energy:

Step 2 — Radius in the magnetic field.

Entering $\vec B$ normally, the magnetic force supplies the centripetal force:

Step 3 — Eliminate $v$.

With $m$, $q$ and $B$ all fixed, the only variable is $V$:

Step 4 — Apply the required change. …

Step 1 — Derive the escape speed (why the formula).

To just escape, the body's total mechanical energy must be zero (it reaches infinity with zero speed):

Note it depends on $r$, the distance from the centre of the Earth, not on the height alone.

Step 2 — Escape speed at the Earth's surface.

Here $r = R$:

Step 3 — Escape speed at the space station.

The station is at height $h = R$, so its distance from the centre is

Step 4 — Take the ratio. …