Physics 9702/42 — February/March 2019

Background Concept

Gravitational potential $V$ at a point is an energy-per-unit-mass quantity.

It is defined using a reference level: for isolated masses we take $V = 0$ at infinity. The idea is to compare how much energy per kilogram is associated with being at a particular position in a gravitational field compared with being infinitely far away.

Understanding the Question

You are asked for the definition (not a formula). The definition must include:

• “work done”
• “per unit mass”
• “from infinity to the point”
• a condition such as “no change in kinetic energy” (or “slowly”).

Approach

Write the standard Cambridge definition. Make sure it is per unit mass (that distinguishes potential from potential energy) and includes infinity as the reference point.

Step-by-Step Reasoning

1. Take a small test mass so it does not significantly change the field.
2. Move it from infinity (where $V=0$ by definition) to the point.
3. If it is moved slowly / with no change in kinetic energy, any work done corresponds to a change in potential energy only.
4. Divide that work by the mass to obtain “per unit mass”, i.e. gravitational potential.

Key Takeaways

• Gravitational potential is work done per unit mass.
• For isolated masses, $V=0$ at infinity.
• The “no change in kinetic energy” condition prevents the work going into kinetic energy.

Common Mistakes

• Giving gravitational potential energy instead (work done, not per unit mass).
• Missing “from infinity”.
• Not stating the condition (slowly / no change in KE).

Things to Be Careful About

• Use “external agent” (or equivalent wording) so the sign convention is clear.
• “Potential at a point” is a scalar and has units $\text{J kg}^{-1}$ (do not write $\text{J}$).

Background Concept

Gravitational potential $V$ is defined as the work done per unit mass by an external agent in bringing a test mass from infinity to the point with no change in kinetic energy.

For an isolated mass, we choose the reference level:

• $V=0$ at infinity.

Also, the gravitational force is attractive, so if you release a mass it accelerates towards the planet and gains kinetic energy: the field is doing positive work on it.

Understanding the Question

You must explain why $V$ is negative near an isolated mass, using your definition. This is essentially a sign argument:

• what is the sign of the external work needed to bring the mass in slowly?

Approach

Start from the definition with $V(\infty)=0$. Decide whether the external work needed is positive or negative when moving inward. Conclude the sign of $V$.

Step-by-Step Reasoning

1. Set the reference: for an isolated mass, define $V=0$ at infinity.
2. Consider moving a $1\ \text{kg}$ test mass from infinity to a point at distance $r$ from the mass.
3. The gravitational force acts towards the mass, in the same direction as the displacement when moving inward, so the gravitational force does positive work on the mass.
4. If the mass were not controlled, it would gain kinetic energy.
5. But the definition of potential requires “no change in kinetic energy” (moved slowly). To prevent it speeding up, an external agent must apply a force opposite the motion and therefore do negative work overall (it must remove the energy that gravity would otherwise supply).
6. Therefore the work done by the external agent per unit mass is negative, so $V<0$ at any finite $r$.

(Equivalently, the standard result for an isolated mass is $V=-\dfrac{GM}{r}$, which is always negative because $G$, $M$, and $r$ are positive.)

Key Takeaways

• Zero potential is chosen at infinity for isolated masses.
• Gravity does positive work as objects fall in.
• Therefore external work required to bring a mass in slowly is negative, so $V$ is negative.

Common Mistakes

• Saying “potential is negative because potential energy is negative” without linking to the definition/reference level.
• Claiming the external agent does positive work when moving inward (sign error).
• Forgetting that $V=0$ is defined at infinity (not at the surface).

Things to Be Careful About

• State clearly whose work you are talking about: the definition uses work done by an external agent.
• Make explicit the “no change in kinetic energy / slowly” condition; otherwise the work could be split between potential and kinetic energy.

Background Concept

A satellite in a circular orbit moves at constant speed but continuously changes direction, so it has centripetal acceleration.

For a circular orbit of radius $r$:

• centripetal force required is $\dfrac{mv^2}{r}$
• gravitational attraction provides this force: $\dfrac{GMm}{r^2}$

Equating them gives the standard orbital-speed relation:

$$\frac{GMm}{r^2} = \frac{mv^2}{r} \Rightarrow v = \sqrt{\frac{GM}{r}}$$

Understanding the Question

You are given planet mass $M$, planet radius $R$, and the satellite’s height above the surface $h$. You must show that the satellite’s orbital speed is $7.4 \times 10^3\ \text{m s}^{-1}$.

Key point: the orbital radius is measured from the planet’s centre, so $r = R + h$.

Approach

1. Compute $r = R + h$.
2. Use $v = \sqrt{GM/r}$.
3. Substitute values and round to match the given result.

Step-by-Step Reasoning

1. Find the distance from the centre:

$$r = 6.40\times 10^6 + 9.00\times 10^5 = 7.30\times 10^6\ \text{m}$$

1. Equate forces for circular motion:

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

Cancel $m$ (speed does not depend on the satellite’s mass) and rearrange:

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

1. Substitute:

$$GM = (6.67\times 10^{-11})(6.00\times 10^{24}) \approx 4.00\times 10^{14}$$ $$\frac{GM}{r} \approx \frac{4.00\times 10^{14}}{7.30\times 10^6} \approx 5.48\times 10^7$$ $$v = \sqrt{5.48\times 10^7} \approx 7.4\times 10^3\ \text{m s}^{-1}$$

Key Takeaways

• For circular orbits: $v = \sqrt{GM/r}$.
• Use $r$ from the centre, not the height above the surface.
• Satellite mass cancels.

Common Mistakes

• Using $r = 9.00\times 10^5\ \text{m}$ instead of $r = R + h$.
• Forgetting to take the square root.
• Using $g = 9.81\ \text{m s}^{-2}$ (only valid near Earth’s surface, not generally here).

Things to Be Careful About

• Keep powers of ten consistent when combining $R$ and $h$.
• Quote the final speed with sensible significant figures (here $7.4 \times 10^3$ matches the target).

Background Concept

Gravitational potential energy $E_p$ of a mass $m$ in the gravitational field of an isolated spherical mass $M$ (treatable as a point mass) is

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

where $r$ is the distance from the centre of mass and the zero of potential energy is taken at infinity.

The negative sign arises because energy must be supplied to move a mass from $r$ out to infinity against the attractive force.

Understanding the Question

You are asked for the satellite’s gravitational potential energy when it is in its circular orbit at height $h$ above the planet. So you need:

• $M$ (given)
• $m$ (given)
• $r = R + h$ (must be calculated)

Approach

1. Find orbital radius $r$ from the planet’s centre.
2. Use $E_p = -GMm/r$.
3. Substitute values and keep the negative sign.

Step-by-Step Reasoning

1. Distance from centre:

$$r = R + h = 6.40\times 10^6 + 9.00\times 10^5 = 7.30\times 10^6\ \text{m}$$

1. Apply the potential energy formula:

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

1. Substitute:

$$E_p = -\frac{(6.67\times 10^{-11})(6.00\times 10^{24})(340)}{7.30\times 10^6}$$

Calculate (keeping powers of ten grouped helps):

• $GM \approx 4.00\times 10^{14}$
• $GMm \approx (4.00\times 10^{14})(340) = 1.36\times 10^{17}$
• divide by $7.30\times 10^6$ gives $\approx 1.86\times 10^{10}$
  So

$$E_p \approx -1.86\times 10^{10}\ \text{J}$$

Key Takeaways

• Use $E_p = -GMm/r$ with $E_p=0$ at infinity.
• Always use $r$ from the centre.
• The potential energy in an attractive inverse-square field is negative.

Common Mistakes

• Using $r=h$ instead of $r=R+h$.
• Omitting the minus sign.
• Confusing potential energy ($\text{J}$) with potential ($\text{J kg}^{-1}$).

Things to Be Careful About

• Keep 3 significant figures if your intermediate values support it; do not over-round early.
• Ensure the unit is $\text{J}$ and not $\text{J kg}^{-1}$ for this part.