Physics 9702/41 — October/November 2012

Background Concept

For uniform circular motion of radius $r$ and speed $v$, the required centripetal force is

$$F_c = \frac{mv^2}{r}$$

In a circular orbit around an isolated planet (mass $M$), the only significant force on the satellite (mass $m$) is gravity, with magnitude

$$F_g = \frac{GMm}{r^2}$$

For a stable circular orbit, the gravitational force acts towards the centre and exactly supplies the centripetal force.

Understanding the Question

You must show (derive) the given expression for the satellite’s kinetic energy in a circular orbit of radius $r$ around mass $M$. The key idea is: set gravitational force equal to centripetal force, find $v^2$, then use $E_K = \tfrac{1}{2}mv^2$.

Approach

1. Write down $F_g$ and $F_c$.
2. Equate them because gravity is the centripetal force in an orbit.
3. Rearrange to get $v^2$ in terms of $GM$ and $r$.
4. Substitute into $E_K = \tfrac{1}{2}mv^2$.

Step-by-Step Reasoning

1. Gravitational force on the satellite:

$$F_g = \frac{GMm}{r^2}$$

1. Centripetal force needed for circular motion at speed $v$:

$$F_c = \frac{mv^2}{r}$$

1. In orbit, the only force is gravity, directed towards the centre, so

$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$

Cancel $m$ and rearrange:

$$v^2 = \frac{GM}{r}$$

1. Kinetic energy is

$$E_K = \frac{1}{2}mv^2$$

Substitute $v^2$:

$$E_K = \frac{1}{2}m\left(\frac{GM}{r}\right) = \frac{GMm}{2r}$$

Key Takeaways

• In a circular orbit, gravity supplies the centripetal force.
• Orbital speed satisfies $v^2 = GM/r$.
• Therefore $E_K$ varies as $1/r$ for circular orbits.

Common Mistakes

• Using $F_c = mr\omega^2$ but then mixing $\omega$ and $v$ incorrectly.
• Forgetting that centripetal force is not an extra force: it is provided by gravity here.
• Algebra slip: losing a factor of $r$ when equating $GMm/r^2$ to $mv^2/r$.

Things to Be Careful About

• This result is for a circular orbit (constant $r$ and $v$).
• $r$ is the orbital radius measured from the planet’s centre.
• Cancelling $m$ is valid because $m \neq 0$ and appears on both sides.

Background Concept

For a satellite in a circular orbit around a planet (mass $M$), two key energy expressions apply:

1. From circular motion and gravitation,

$$E_K = \frac{GMm}{2r}$$

1. Taking gravitational potential energy to be zero at infinity,

$$U = -\frac{GMm}{r}$$

When the orbit radius changes from $r_1$ to $r_2$, the changes are

$$\Delta E_K = E_{K2} - E_{K1}, \qquad \Delta U = U_2 - U_1$$

Signs matter: if $r$ decreases, $U$ becomes more negative, so $\Delta U$ is negative.

Understanding the Question

You are given $GM$ for the planet and the satellite mass $m$. The satellite moves from an orbit of radius $r_1 = 7.34 \times 10^6\ \text{m}$ to a slightly smaller orbit $r_2 = 7.30 \times 10^6\ \text{m}$. You must calculate the numerical change in:

• kinetic energy $\Delta E_K$,
• gravitational potential energy $\Delta U$.

Because the two radii are close, it is efficient (and reduces rounding error) to use the difference in $1/r$ rather than calculating two large energies and subtracting.

Approach

1. Compute the constant product $GMm$.
2. Use

$$\Delta E_K = \frac{GMm}{2}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$$

1. Use

$$\Delta U = -GMm\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$$

1. Check signs: since $r_2 < r_1$, the bracket is positive, so $\Delta E_K$ positive and $\Delta U$ negative.

Step-by-Step Reasoning

1. Calculate $GMm$:

$$GMm = (4.00 \times 10^{14})(620) = 2.48 \times 10^{17}$$

(Units are joules when divided by $r$ because $GM$ has units $\text{m}^3\ \text{s}^{-2}$, so $GMm/r$ has units $\text{kg m}^2\ \text{s}^{-2} = \text{J}$.)

1. Compute the difference in reciprocal radii:

$$\frac{1}{r_2}-\frac{1}{r_1} = \frac{1}{7.30\times 10^6}-\frac{1}{7.34\times 10^6} = 7.46\times 10^{-10}\ \text{m}^{-1}$$

1. Kinetic energy change:

$$\Delta E_K = \frac{GMm}{2}\left(\frac{1}{r_2}-\frac{1}{r_1}\right) = \frac{2.48\times 10^{17}}{2}(7.46\times 10^{-10}) = 9.25\times 10^7\ \text{J}$$

Positive means the satellite’s kinetic energy increases.

1. Potential energy change:

$$\Delta U = -GMm\left(\frac{1}{r_2}-\frac{1}{r_1}\right) = -(2.48\times 10^{17})(7.46\times 10^{-10}) = -1.85\times 10^8\ \text{J}$$

Negative means the satellite loses gravitational potential energy (it is closer to the planet, so more tightly bound).

1. Quick consistency check (useful in exams): for circular orbits,

$$U = -2E_K$$

So changes should satisfy approximately $\Delta U \approx -2\Delta E_K$, which matches here: $-2(9.25\times 10^7) \approx -1.85\times 10^8$.

Key Takeaways

• For circular orbits: $E_K \propto 1/r$ and $U \propto -1/r$.
• When $r$ decreases: $E_K$ increases (positive change) and $U$ decreases (negative change).
• Using $\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$ is a neat way to avoid subtracting close large numbers.

Common Mistakes

• Forgetting the minus sign in $U = -GMm/r$.
• Reporting only magnitudes and not the sign of the change.
• Using $\Delta U = GMm(1/r_2 - 1/r_1)$ (wrong sign).
• Rounding $E_{K1}$ and $E_{K2}$ too aggressively before subtracting, giving a poor $\Delta E_K$.

Things to Be Careful About

• Keep $r$ in metres.
• $GM$ is given; do not try to insert a value of $G$ separately.
• Quote final answers to appropriate significant figures and include units ($\text{J}$).

Background Concept

For a mass $m$ moving at speed $v$, kinetic energy is

$$E_K = \frac{1}{2}mv^2$$

So if $E_K$ increases (with constant $m$), then $v^2$ and hence $v$ must increase.

For a circular gravitational orbit, equating gravity to centripetal force gives

$$v^2 = \frac{GM}{r}$$

which also shows directly that decreasing $r$ increases $v$.

Understanding the Question

You must use your calculated changes in kinetic energy and potential energy to decide what happens to the satellite’s linear speed when the orbit radius decreases from $7.34\times 10^6\ \text{m}$ to $7.30\times 10^6\ \text{m}$.

Approach

Use the sign of $\Delta E_K$ from (ii). If $\Delta E_K > 0$, then the speed must have increased. You can also support it by quoting $v^2 = GM/r$.

Step-by-Step Reasoning

1. From (ii), the change in kinetic energy is positive ($\Delta E_K \approx +9.3\times 10^7\ \text{J}$). Therefore $E_K$ is larger in the smaller-radius orbit.

2. Since

$$E_K = \frac{1}{2}mv^2$$

and $m$ is constant, an increase in $E_K$ implies an increase in $v^2$, so $v$ increases.

1. This matches the orbit relation:

$$v^2 = \frac{GM}{r}$$

If $r$ decreases, the right-hand side increases, so $v$ must increase.

1. Note how the energies fit together physically: the satellite loses gravitational potential energy (becomes more negative) and gains some kinetic energy; overall, the total mechanical energy becomes more negative (energy is dissipated by resistive forces), but the speed can still increase.

Key Takeaways

• Positive change in kinetic energy means speed increases.
• For circular orbits, smaller radius means higher orbital speed ($v \propto r^{-1/2}$).
• Decreasing radius can occur even with resistive forces because total energy decreases while kinetic energy can increase.

Common Mistakes

• Thinking “resistive forces always slow it down” and concluding speed decreases; in orbital decay the speed in the new lower circular orbit is actually higher.
• Using $E_K \propto r$ instead of $E_K \propto 1/r$.
• Ignoring the sign of the change and comparing magnitudes only.

Things to Be Careful About

• The conclusion applies to circular orbits at the stated radii (i.e. comparing speeds of two circular orbits). During the transition the motion need not be circular at every instant.
• Make the reasoning explicit: link $\Delta E_K$ to $v$ via $E_K = \tfrac{1}{2}mv^2$ (or quote $v^2 = GM/r$).