Gravitational Fields: Question 3

Worked solution

Part (a): Defining gravitational potential and explaining its sign

Definition: the gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point.

Why it is negative: gravitational forces are always attractive. To move a test mass away from Virellon, out towards infinity, an external agent must do positive work against this attraction. Equivalently, if the test mass instead moves inward, from infinity to a point near Virellon, the gravitational force itself does positive work on it, so the work that would need to be done by an external agent to achieve this same inward motion (quasi-statically, without gaining kinetic energy) is negative. Since gravitational potential is defined as the work done per unit mass bringing the mass in from infinity, and infinity is taken as the zero of potential, the potential at every finite distance from Virellon is therefore negative.

Part (b): Gravitational potential at $8.00\times10^6\text{ m}$

For a point mass, $\phi = -GM/r$. With $M=4.20\times10^{25}\text{ kg}$ and $r=8.00\times10^6\text{ m}$: $\phi = -\frac{GM}{r} = -\frac{(6.67\times10^{-11})(4.20\times10^{25})}{8.00\times10^6}$

Numerator: $6.67\times4.20=28.014$, and $10^{-11}\times10^{25}=10^{14}$, so the numerator is $2.8014\times10^{15}$.

$\phi = -\frac{2.8014\times10^{15}}{8.00\times10^6} = -3.50175\times10^8\text{ J kg}^{-1} \approx -3.50\times10^8\text{ J kg}^{-1}$

Check by recomputing differently: $2.8014/8.00=0.35018$, and $10^{15}/10^{6}=10^{9}$, so $\phi=-0.35018\times10^{9}=-3.5018\times10^8\text{ J kg}^{-1}$. Both routes agree.

Part (c): Change in gravitational potential energy

The gravitational potential energy of the probe at a point is $E_p=m\phi$.

At $r_1=8.00\times10^6\text{ m}$, using $\phi_1=-3.50175\times10^8\text{ J kg}^{-1}$ from (b): $E_{p1} = m\phi_1 = 250\times(-3.50175\times10^8) = -8.754\times10^{10}\text{ J}$

At $r_2=1.60\times10^7\text{ m}$ (exactly twice $r_1$), since $\phi\propto1/r$, the potential is halved: $\phi_2 = \frac{\phi_1}{2} = -1.75088\times10^8\text{ J kg}^{-1}$

Check by recalculating directly: $\phi_2=-GM/r_2=-2.8014\times10^{15}/(1.60\times10^7)=-1.75088\times10^8\text{ J kg}^{-1}$, matching the halved value exactly.

$E_{p2} = m\phi_2 = 250\times(-1.75088\times10^8) = -4.377\times10^{10}\text{ J}$

The change in potential energy is: $\Delta E_p = E_{p2}-E_{p1} = (-4.377\times10^{10})-(-8.754\times10^{10}) = +4.377\times10^{10}\text{ J} \approx +4.38\times10^{10}\text{ J}$

Check using the shortcut formula $\Delta E_p = GMm\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$: since $r_2=2r_1$, this is $GMm\left(\dfrac{1}{r_1}-\dfrac{1}{2r_1}\right)=\dfrac{GMm}{2r_1}=\dfrac{2.8014\times10^{15}\times250}{2\times8.00\times10^6}=\dfrac{7.0035\times10^{17}}{1.60\times10^7}=4.377\times10^{10}\text{ J}$. This matches, confirming $\Delta E_p\approx+4.38\times10^{10}\text{ J}$.

Since $\Delta E_p$ is positive, the potential energy increases as the probe moves away from Virellon. Consistent with gravitational potential energy always increasing as distance from an attracting mass increases.

Part (d): Comparing the shapes of the $g$–$r$ and $\phi$–$r$ graphs

Similarity: for both quantities, the magnitude decreases continuously as $r$ increases, and both tend towards zero as $r\to\infty$ (neither graph ever reaches zero at a finite $r$).

Difference: the field strength $g=GM/r^2$ is always positive and falls off very steeply, as $1/r^2$, so it drops rapidly towards zero at large $r$. The potential $\phi=-GM/r$ is always negative and falls off much more gradually, as $1/r$ (i.e. its magnitude decreases more slowly than that of $g$), rising smoothly from large negative values towards zero as $r$ increases.

Final answers

• (a) Gravitational potential is the work done per unit mass bringing a test mass from infinity to the point; it is negative because gravity is attractive, so bringing a mass in from infinity involves negative work by an external agent (equivalently, the field itself does positive work pulling the mass in).
• (b) $\phi = \boxed{-3.50\times10^8}\text{ J kg}^{-1}$
• (c) $\Delta E_p = \boxed{+4.38\times10^{10}}\text{ J}$, an increase
• (d) Both magnitudes fall towards zero as $r\to\infty$; but $g$ is always positive and falls off as $1/r^2$ (steeply), while $\phi$ is always negative and falls off as $1/r$ (more gradually).