Issue 082 - Spaceflight - Kinetic energy
How much energy must the atmosphere remove from Swift?
NASA's November 2 forecast expected its roughly 3,200-pound Swift Observatory to reenter on November 4, 2026, after almost 22 years in orbit.
Space and astronomyAbout 1 minute
Sources checked November 3, 2026. Figures and circumstances may have changed.
The problem
Estimate the kinetic energy in joules of a roughly 1.5-metric-ton spacecraft traveling in low Earth orbit shortly before reentry.
Choose an orbital speed, then compare the energy with gasoline or a highway-speed car. The guess unit is joules of orbital kinetic energy, not power or a prediction of ground impact.
Optionally compare its potential energy above Earth's surface at 275 km altitude, or estimate average power if most of the energy is removed over a few minutes.
Fermi estimates target scale rather than precision. The Fermi reference collects useful facts and equations.
Grounding facts
NASA's November 2 update gives a mass of roughly 3,200 pounds and a November 4 reentry forecast with substantial timing uncertainty. It expects most of Swift to burn up, with some components surviving. This article records that forecast, not a confirmed reentry date or location.
NASA reported that Swift reached about 275 km altitude on October 20, when increased atmospheric drag forced it into safe mode. That is a dated altitude anchor, not its measured altitude at publication.
ESA gives typical low-Earth-orbit speed as about 7.8 km/s; NASA's orbital-debris guidance gives 7 to 8 km/s. By contrast, 30 km/s is a useful peg for Earth's orbit around the Sun. It is not Swift's speed relative to Earth's atmosphere.
DOE's gasoline energy-content table corresponds to roughly 32 to 35 million joules per liter, depending on blend and heating-value convention. Use 34 million J/L for this comparison. A one-ton car at 30 m/s is an illustrative comparison chosen here, not an average vehicle specification.
After checking sources
Checked answer and calculation
mass ~= 1.5 x 10^3 kg
low-Earth-orbit speed ~= 7.8 x 10^3 m/s
kinetic energy = (1/2) m v^2
~= 0.5 x 1.5 x 10^3 x (7.8 x 10^3)^2
~= 4.6 x 10^10 J
~= 45 billion joules, rounded
one-ton car at 30 m/s
= 0.5 x 1,000 x 30^2
= 4.5 x 10^5 J
car-energy equivalents
~= 4.5 x 10^10 / 4.5 x 10^5
~= 100,000 cars
gasoline-energy equivalent
~= 4.5 x 10^10 / 3.4 x 10^7 J/L
~= 1,300 litersA modest satellite mass therefore carries tens of billions of joules, far beyond a single car crash. These are energy-content comparisons, not claims that reentry acts like 100,000 simultaneous crashes or a concentrated ground explosion. Energy is distributed along the trajectory among shocked air, radiation, spacecraft heating and breakup; surviving debris can retain some energy.
Using 7 to 8 km/s and a mass near 1,450 to 1,500 kg puts the kinetic-energy scale around 35 to 50 billion joules. The rounded checked target is 45 billion J. Small differences between Earth-centered speed and speed relative to rotating air do not alter the Fermi scale.
potential energy above the surface, approximate
= m g h
~= 1.5 x 10^3 x 10 x 2.75 x 10^5
~= 4.1 x 10^9 J
kinetic / altitude potential energy
~= 4.5 x 10^10 / 4.1 x 10^9
~= 11The constant-g potential-energy approximation is adequate at this altitude. Accounting for gravity decreasing with height gives about 3.9 billion J, rather than changing the order of magnitude. This is the potential-energy difference from 275 km to the surface, not the magnitude of gravitational binding energy relative to infinity.
From that altitude, roughly 50 billion joules of mechanical energy must ultimately be redistributed to end near rest at the surface, counting both kinetic energy and the potential-energy drop. Nearly all orbital kinetic energy is lost before surviving pieces reach the ground. The detailed split and timing require a reentry model.
illustrative intense phase = 5 minutes = 300 s
average power from orbital kinetic energy
~= 4.5 x 10^10 J / 300 s
~= 1.5 x 10^8 W
~= 150 MW
for a 2-to-10-minute assumed interval
~= 75 to 380 MWThat is comparable in rate to a substantial power-plant unit, but only for minutes, and it is dissipated energy rather than electricity generation. The interval is an assumption, not a measured heating history for Swift.
Matt's 30,000 m/s peg is about four times too fast here; squaring it makes the kinetic energy about 15 times too high. Separately, his stated mgh inputs multiply to 4.2 billion J, not 110 billion J. The potential-energy arithmetic error is independent of the speed mistake.
Before checking sources
Matt's original estimate
This is Matt's original, unverified estimate, preserved before the source check.
All of my calculations depend on the assumption that the orbital velocity was about 30,000 m/s. I don't know why that number is in my head, but that's what I used for all calculations.
KE = 1/2 * m * v^2, so that's 1/2 * (1.5 * 10^3 kg) * (3 * 10^4 m/s)^2 which comes to about 6.8 * 10^11 J
comparing that to a highway car that weighs about a ton and is moving at about 60 mph, that's 1/2 * (10^3 kg) * (30 m/s)^2 = about 4.5 * 10^5 J per car, so it would take about 1.5 million cars moving at highway speed to equal the same kinetic energy.
comparing that to the energy in a liter of gasoline, that's 4.5 * 10^7 J / L, so that's about 1.5 * 10^4 L of fuel to equal the same amount of energy.
then looking at the PE at 275 km above earth's surface, PE = m * g * h so that's (1.5 * 10^3 kg) * (10 m/s^2) * (2.8 * 10^5 m) = about 1.1 * 10^11 J, so its kinetic energy is about 6x larger than it's potential energy at that elevation/altitude.
Reasoning score
Matt's reasoning score: 40 / 100
Useful facts: 0/30. The decisive orbital-speed peg confused Earth-centered motion with Earth's solar orbit. The gasoline peg was also high per liter; mass, gravity and highway speed were reasonable.
Reasoning approach: 30/30. Kinetic energy, mgh for the modest altitude difference, and ratios to familiar energy scales were appropriate models.
Math: 0/10. The kinetic-energy arithmetic followed the assumptions, but the potential-energy product was about 26 times too large and materially changed the KE-to-PE comparison.
Final estimate: 10/30. The 680-billion-joule estimate is about 15 times the checked 45-billion-joule target: outside one order of magnitude but within two.
Post-check reflection
Matt's reflection
My memory peg for objects in orbit around us messed me up - I was actually thinking of the velocity of objects in orbit around the sun. Threw me off on my answers by an order of magnitude too high.
Otherwise this one was pretty straightforward.
Recommended memory peg
Low-Earth-orbit speed ~= 8 km/s; Earth's speed around the Sun ~= 30 km/s. Choose the reference frame before using KE = (1/2)mv^2. At 8 km/s, each kilogram carries roughly 32 million joules of kinetic energy.