Gravitational Potential
Q1. Define gravitational potential at a point and state its unit. 2 marks
Q2. Earth's mass is 5.97 × 1024 kg and radius is 6.37 × 106 m. Calculate the gravitational potential:
(a) at Earth's surface 1 mark
(b) at 1000 km altitude 1 mark
(c) at infinity 1 mark
Q3. With outward radius positive, explain the relationship between signed radial field component $g_r$, field magnitude $|\vec g|$ and gravitational potential $V$. 2 marks
Q4. A mass of 2.0 kg moves from a point where V = −50 MJ/kg to a point where V = −30 MJ/kg.
(a) Calculate the change in gravitational potential energy. 1 mark
(b) Calculate the work done by gravity and interpret its sign. 1 mark
(c) Calculate the work done by an external force. 1 mark
Q5. The gravitational potential at distance r from a point mass M is V = −GM/r.
(a) Taking outward $r$ as positive, show that $g_r = -dV/dr$. 2 marks
(b) Sketch $V$ versus $r$ and field magnitude $|\vec g|$ versus $r$ on separate labelled axes. 2 marks
Q6. Consider the gravitational potential due to Earth.
(a) Explain why equipotential surfaces around a spherical mass are spherical. 2 marks
(b) Describe how the spacing of equipotential surfaces changes with distance from the mass. 2 marks
(c) A satellite moves along an equipotential surface. Explain why no work is done by gravity. 2 marks
Answer Key
Q1. Gravitational potential at a point is the work done per unit mass in bringing a test mass from infinity to that point. Unit: J/kg.
Q2. (a) V = −6.67×10−11×5.97×1024/6.37×106 = −6.25×107 J/kg = −62.5 MJ/kg (b) V = −6.67×10−11×5.97×1024/7.37×106 = −54.0 MJ/kg (c) V = 0
Q3. $g_r=-dV/dr$. For a point mass, $dV/dr=GM/r^2=|\vec g|$. Thus $g_r=-GM/r^2$: the negative component means inward, while the magnitude is positive.
Q4. (a) $\Delta U=m\Delta V=2.0\times20\times10^6=+4.0\times10^7$ J (b) $W_g=-\Delta U=-4.0\times10^7$ J; negative because gravity opposes the outward displacement (c) for a quasistatic transfer, $W_{\rm ext}=+\Delta U=+4.0\times10^7$ J.
Q5. (a) $dV/dr=d(-GM/r)/dr=+GM/r^2$. Therefore $g_r=-dV/dr=-GM/r^2$ for outward-positive $r$, while $|\vec g|=GM/r^2$. (b) $V$ versus $r$: negative curve increasing toward zero; $|\vec g|$ versus $r$: positive curve decreasing as $1/r^2$. Axes must be separate because the quantities have different units.
Q6. (a) $V$ depends only on $r$ for a spherical mass, so all points at the same $r$ form a spherical equipotential. (b) Closer to the mass, $|\vec g|$ and the potential gradient are larger, so equal potential intervals are closer together; farther away they are wider. (c) Along an equipotential, $\Delta V=0$, so $W_g=-m\Delta V=0$ (and quasistatic $W_{\rm ext}=m\Delta V=0$).
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