Lecture 6 Solutions: Newton’s Laws & Gravity
Practice Problem Solutions
Student note: These are model solutions written to show every step, units, and checks. I do not expect your work to be this detailed; shorter solutions are fine if your setup, units, and reasoning are correct.
HW note: Homework uses only a subset of these problems; the full set is included for future study and self checking.
Core Problems
Problem 1: Newton’s First Law (Probe in Space)
Restatement: Describe the motion of a probe coasting at 50 km/s through interstellar space with empty fuel tanks and negligible gravitational effects.
Key insight: Newton’s First Law — an object in motion stays in motion at constant velocity unless acted upon by a net force.
Answer: The probe continues at 50 km/s in a straight line, forever. With no fuel (no thrust force) and negligible gravity, there is no net force. Newton’s First Law says an object in motion remains in uniform motion unless a force acts on it. There is no friction in space, no air resistance, and no drag — the probe does not “slow down” or “run out of steam.” A million years later, it is still moving at 50 km/s in the same direction.
Common misconception: “It will eventually slow down and stop.” This is wrong — that intuition comes from Earth, where friction is always present. In space, there is nothing to provide a decelerating force.
What you should learn from this: Newton’s First Law is the default state of the universe. Forces cause changes in motion, not motion itself.
Problem 2: Newton’s Second Law (Car Acceleration)
Restatement: Find the net force on a 1000 kg car accelerating at 3 m/s².
Given:
- Mass: \(m = 1000\) kg
- Acceleration: \(a = 3 \text{ m/s}^2\)
Find: Net force \(F\) in newtons.
Equation: \[F = ma\]
Solution: \[F = (1000 \text{ kg}) \times (3 \text{ m/s}^2) = 3000 \text{ kg·m/s}^2 = 3000 \text{ N}\]
Unit check: \(\text{kg} \times \text{m/s}^2 = \text{kg·m/s}^2 = \text{N}\) (the definition of a newton) ✓
Sanity check: A typical car engine might produce a few thousand newtons of net force during modest acceleration — 3000 N is reasonable. ✓
Answer: \(F = 3000\) N (3 kN).
What you should learn from this: Newton’s Second Law (\(F = ma\)) is the central equation of mechanics. It says force, mass, and acceleration are linked: more mass requires more force for the same acceleration.
Problem 3: Inverse-Square Law (Gravitational Acceleration)
Restatement: At what distance from Earth’s center is gravitational acceleration 1/4 as strong as at the surface?
Given:
- Earth’s surface radius: \(R_\oplus = 6370\) km
- Gravitational acceleration at surface: \(g\)
- Required: \(g_{\text{new}} = g/4\)
Find: Distance \(r\) from Earth’s center.
Equation: Gravitational acceleration: \(g \propto 1/r^2\)
\[\frac{g_{\text{new}}}{g_{\text{surface}}} = \left(\frac{R_\oplus}{r}\right)^2\]
Solution:
\[\frac{1}{4} = \left(\frac{R_\oplus}{r}\right)^2\]
Take the square root of both sides:
\[\frac{1}{2} = \frac{R_\oplus}{r}\]
\[r = 2 R_\oplus = 2 \times 6370 \text{ km} = 12{,}740 \text{ km}\]
Unit check: km × (dimensionless) = km ✓
Sanity check: The inverse-square law says gravity drops as \(1/r^2\). To get 1/4 the gravity, we need \(r^2\) to be 4 times larger, so \(r\) must be 2 times larger. Double the distance → quarter the gravity. ✓
Answer: \(r = 2R_\oplus = 12{,}740\) km from Earth’s center (i.e., 6370 km above the surface — one Earth radius of altitude).
What you should learn from this: The inverse-square law is a powerful scaling tool. You don’t need to know \(G\) or \(M_\oplus\) — just the ratio. Double the distance → quarter the force.
Problem 4: Weight vs. Mass
Restatement: Calculate your gravitational force (weight) on Earth and on the Moon. Does your mass change?
Given:
- Mass: \(m = 70\) kg
- Earth: \(g_{\text{Earth}} = 9.8 \text{ m/s}^2\)
- Moon: \(g_{\text{Moon}} = 1.6 \text{ m/s}^2\)
Find: Weight (gravitational force) on each body.
Equation: \[W = mg\]
Solution:
On Earth: \[W_{\text{Earth}} = (70 \text{ kg})(9.8 \text{ m/s}^2) = 686 \text{ N}\]
On the Moon: \[W_{\text{Moon}} = (70 \text{ kg})(1.6 \text{ m/s}^2) = 112 \text{ N}\]
Unit check: \(\text{kg} \times \text{m/s}^2 = \text{N}\) ✓
Does mass change? No. Mass is an intrinsic property — it measures how much matter is in you. It is 70 kg on Earth, on the Moon, and in deep space. Weight is a force — it depends on the local gravitational field. Your mass is always 70 kg; your weight changes depending on where you are.
Sanity check: The Moon’s gravity is about 1/6 of Earth’s (\(1.6/9.8 \approx 0.16\)), so your weight should be about 1/6: \(686/6 \approx 114\) N. Our answer of 112 N is consistent. ✓
Answer: \(W_{\text{Earth}} = 686\) N; \(W_{\text{Moon}} = 112\) N. Mass (70 kg) does not change.
What you should learn from this: Mass and weight are different quantities with different units. Mass (kg) is constant; weight (N) depends on gravity. Astronauts are not “massless” — they are “weightless” (in free fall).
Problem 5: Weightlessness on the ISS
Restatement: Explain why astronauts appear weightless on the ISS even though gravity is ~90% as strong as on the surface.
Key insight: Weightlessness is not about the absence of gravity — it is about free fall.
Answer: The ISS orbits at about 400 km altitude, where gravity is roughly 90% of its surface value (\(g \approx 8.7 \text{ m/s}^2\) vs. \(9.8 \text{ m/s}^2\)). Astronauts are not beyond gravity’s reach.
The reason they appear weightless is that the ISS and everything inside it — astronauts, equipment, water droplets — are all falling toward Earth at the same rate. The station is in free fall, but it’s also moving sideways fast enough (~7.7 km/s) that it keeps missing Earth. Since astronauts and the station fall together, there is no contact force between them — no floor pushing up, no seat pressing against them. This absence of contact forces is what we perceive as “weightlessness.”
An analogy: if you were in an elevator and the cable snapped, you and the elevator would fall together. For those brief seconds, you’d feel weightless — not because gravity disappeared, but because nothing is pushing back against you.
Common misconception: “There’s no gravity in space.” Gravity extends to infinity (though it weakens with distance). At ISS altitude, it’s barely reduced.
What you should learn from this: “Weightless” means “in free fall,” not “beyond gravity.” This concept is essential for understanding orbits: orbiting is falling, with enough sideways velocity to keep missing.
Problem 6: Planet Mass from Moon Orbit
Restatement: Calculate a planet’s mass from its moon’s orbital parameters.
Given:
- Moon orbital radius: \(a = 500{,}000\) km
- Moon orbital period: \(P = 10\) days
- \(G = 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}\)
- Assume circular orbit and \(m_{\text{moon}} \ll M_{\text{planet}}\)
Find: Planet mass \(M\) in kg.
Equation: \[M = \frac{4\pi^2 a^3}{GP^2}\]
Solution:
Step 1 — Convert to SI units:
- \(a = 500{,}000 \text{ km} \times \frac{1000 \text{ m}}{1 \text{ km}} = 5.00 \times 10^8 \text{ m}\)
- \(P = 10 \text{ days} \times \frac{86{,}400 \text{ s}}{1 \text{ day}} = 8.64 \times 10^5 \text{ s}\)
Step 2 — Compute the numerator (\(4\pi^2 a^3\)):
\[a^3 = (5.00 \times 10^8 \text{ m})^3 = 1.25 \times 10^{26} \text{ m}^3\]
\[4\pi^2 \times 1.25 \times 10^{26} \text{ m}^3 = 39.48 \times 1.25 \times 10^{26} \text{ m}^3 = 4.93 \times 10^{27} \text{ m}^3\]
Step 3 — Compute the denominator (\(GP^2\)):
\[P^2 = (8.64 \times 10^5 \text{ s})^2 = 7.46 \times 10^{11} \text{ s}^2\]
\[GP^2 = (6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}) \times (7.46 \times 10^{11} \text{ s}^2) = 4.98 \times 10^{1} \text{ m}^3 \text{ kg}^{-1}\]
Check units: \(\text{m}^3 \text{ kg}^{-1} \text{ s}^{-2} \times \text{s}^2 = \text{m}^3 \text{ kg}^{-1}\) ✓
Step 4 — Divide:
\[M = \frac{4.93 \times 10^{27} \text{ m}^3}{4.98 \times 10^{1} \text{ m}^3 \text{ kg}^{-1}} = 9.9 \times 10^{25} \text{ kg}\]
Check units: \(\frac{\text{m}^3}{\text{m}^3 \text{ kg}^{-1}} = \text{kg}\) ✓
Sanity check: For reference, Earth is \(6 \times 10^{24}\) kg and Neptune is \(1.0 \times 10^{26}\) kg. Our planet at \(9.9 \times 10^{25}\) kg is about 16× Earth’s mass — comparable to Neptune. The orbital parameters (500,000 km, 10 days) are similar to the outer planets’ moon systems, so this is reasonable. ✓
Answer: \(M \approx 9.9 \times 10^{25}\) kg (about 16 Earth masses — similar to an ice giant like Neptune).
What you should learn from this: This is the same technique used to “weigh” any planet or star: observe something orbiting it, measure the orbit, and apply Newton’s form of Kepler III. The template is always: convert to SI → cube \(a\) → square \(P\) → divide.
Challenge Problems
Problem 7: Deriving Kepler III from Newton
Restatement: Starting from centripetal and gravitational force equations, derive \(v = \sqrt{GM/r}\) and then Kepler’s Third Law for circular orbits.
Key insight: Setting gravitational force equal to centripetal force links orbital velocity to mass and distance.
Answer:
Part 1 — Derive \(v = \sqrt{GM/r}\):
For a circular orbit, gravity provides the centripetal force:
\[F_{\text{gravity}} = F_{\text{centripetal}}\]
\[\frac{GMm}{r^2} = \frac{mv^2}{r}\]
The orbiting mass \(m\) cancels (it appears on both sides):
\[\frac{GM}{r^2} = \frac{v^2}{r}\]
Multiply both sides by \(r\):
\[\frac{GM}{r} = v^2\]
\[\boxed{v = \sqrt{\frac{GM}{r}}}\]
Check units: \(\sqrt{\frac{\text{m}^3 \text{ kg}^{-1} \text{ s}^{-2} \times \text{kg}}{\text{m}}} = \sqrt{\frac{\text{m}^2}{\text{s}^2}} = \text{m/s}\) ✓
Part 2 — Derive Kepler III:
For a circular orbit, \(v = 2\pi r / P\) (circumference divided by period). Substitute:
\[\frac{2\pi r}{P} = \sqrt{\frac{GM}{r}}\]
Square both sides:
\[\frac{4\pi^2 r^2}{P^2} = \frac{GM}{r}\]
Multiply both sides by \(r\):
\[\frac{4\pi^2 r^3}{P^2} = GM\]
Rearrange:
\[\boxed{P^2 = \frac{4\pi^2}{GM} r^3}\]
This is Kepler’s Third Law: \(P^2 \propto r^3\), with the proportionality constant \(4\pi^2/(GM)\) depending on the central mass \(M\).
What you should learn from this: Newton derived Kepler’s empirical law from first principles. Kepler described what happens; Newton explained why. The cancellation of \(m\) is physically important — it means orbital motion doesn’t depend on the orbiting object’s mass.
Problem 8: Alternative Gravity Law
Restatement: If gravity followed \(1/r^3\) instead of \(1/r^2\), would orbits still be ellipses? Would \(P^2 \propto a^3\) hold?
Key insight: The \(1/r^2\) law is special — it uniquely produces closed elliptical orbits.
Answer:
Would orbits be ellipses? No. It is a mathematical result (proved by Newton using calculus) that only the \(1/r^2\) force law produces closed elliptical orbits. A \(1/r^3\) law would produce orbits that are not closed — the path would precess (rotate) so the planet never returns to exactly the same point. The orbits would spiral inward or outward depending on initial conditions.
Would \(P^2 \propto a^3\) hold? No. Following the derivation from Problem 7 but with \(F \propto 1/r^3\):
\[\frac{GM}{r^3} = \frac{v^2}{r} \quad \Rightarrow \quad v^2 = \frac{GM}{r^2} \quad \Rightarrow \quad v = \sqrt{\frac{GM}{r^2}}\]
Using \(v = 2\pi r/P\):
\[\frac{4\pi^2 r^2}{P^2} = \frac{GM}{r^2} \quad \Rightarrow \quad P^2 = \frac{4\pi^2}{GM} r^4\]
So we’d get \(P^2 \propto r^4\), not \(P^2 \propto r^3\).
What you should learn from this: The \(1/r^2\) law isn’t arbitrary — it’s what produces stable, closed orbits and the specific \(P^2 \propto a^3\) relationship. If gravity worked differently, the solar system would look fundamentally different (or might not be stable at all).
Problem 9: Mars’s Mass from Phobos
Restatement: Calculate Mars’s mass from Phobos’s orbital parameters.
Given:
- Phobos orbital radius: \(a = 9376\) km (from Mars’s center)
- Phobos orbital period: \(P = 7.7\) hours
- \(G = 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}\)
- Earth’s mass (for comparison): \(M_\oplus = 6 \times 10^{24}\) kg
Find: Mars’s mass \(M\) in kg.
Equation: \[M = \frac{4\pi^2 a^3}{GP^2}\]
Solution:
Step 1 — Convert to SI units:
- \(a = 9376 \text{ km} \times 1000 \text{ m/km} = 9.376 \times 10^6 \text{ m}\)
- \(P = 7.7 \text{ hr} \times 3600 \text{ s/hr} = 2.772 \times 10^4 \text{ s}\)
Step 2 — Compute the numerator (\(4\pi^2 a^3\)):
\[a^3 = (9.376 \times 10^6 \text{ m})^3 = 8.246 \times 10^{20} \text{ m}^3\]
\[4\pi^2 \times 8.246 \times 10^{20} \text{ m}^3 = 39.48 \times 8.246 \times 10^{20} = 3.256 \times 10^{22} \text{ m}^3\]
Step 3 — Compute the denominator (\(GP^2\)):
\[P^2 = (2.772 \times 10^4 \text{ s})^2 = 7.684 \times 10^8 \text{ s}^2\]
\[GP^2 = (6.674 \times 10^{-11}) \times (7.684 \times 10^8) = 5.128 \times 10^{-2} \text{ m}^3 \text{ kg}^{-1}\]
Check units: \(\text{m}^3 \text{ kg}^{-1} \text{ s}^{-2} \times \text{s}^2 = \text{m}^3 \text{ kg}^{-1}\) ✓
Step 4 — Divide:
\[M = \frac{3.256 \times 10^{22}}{5.128 \times 10^{-2}} = 6.3 \times 10^{23} \text{ kg}\]
Check units: \(\frac{\text{m}^3}{\text{m}^3 \text{ kg}^{-1}} = \text{kg}\) ✓
Comparison to Earth: \[\frac{M_\oplus}{M_{\text{Mars}}} = \frac{6 \times 10^{24}}{6.3 \times 10^{23}} \approx 9.5\]
Earth is about 9.5 times more massive than Mars.
Sanity check: Mars is smaller than Earth (about half the diameter), so it should be significantly less massive. A factor of ~10 is consistent with Mars’s smaller size and lower density. The accepted value is \(6.42 \times 10^{23}\) kg — our answer is within 2%. ✓
Answer: \(M_{\text{Mars}} \approx 6.3 \times 10^{23}\) kg. Earth is about 9.5× more massive.
What you should learn from this: The same technique works for any central body — just plug in the moon’s orbital parameters. Notice this gives nearly the same answer as the practice problem solution in practice-problems.qmd (which uses a different Mars satellite), providing an independent cross-check.
Problem 10: Why Sun Units Simplify Kepler III
Restatement: Explain why \(P^2 = a^3\) works only for the Sun as the central mass, with \(P\) in years and \(a\) in AU.
Key insight: The simplified form absorbs the constants \(4\pi^2/(GM_\odot)\) into the choice of units.
Answer:
The full form of Kepler’s Third Law is:
\[P^2 = \frac{4\pi^2}{GM} a^3\]
When we choose years for \(P\) and AU for \(a\), the constant \(4\pi^2/(GM_\odot)\) happens to equal exactly 1 (by definition — the AU and year are defined relative to Earth’s orbit around the Sun). So for Sun-orbiting objects:
\[P^2 = 1 \times a^3 = a^3\]
For a different central mass (say Jupiter), \(4\pi^2/(GM_{\text{Jupiter}})\) has a different value, so the shorthand breaks. You’d need either:
- The full equation with \(G\) and \(M\) in SI units, or
- A new set of “Jupiter units” (e.g., \(P\) in units of Io’s period, \(a\) in units of Io’s orbit)
The simplification works because Earth’s orbit defines both the AU and the year. For any other system, the constants don’t cancel so neatly.
What you should learn from this: “Natural units” are a powerful trick: by choosing units matched to your system, equations simplify dramatically. But you must know when the simplification applies and what changes when you switch systems.