Exoplanet Transit Detection Calculator

Q: Does a transit tell me the planet’s mass?

No. It gives you the planet’s size relative to its star, and nothing about how heavy it is. A solid iron ball and a ball of fog of the same radius block exactly the same light. Mass comes from watching the star wobble towards and away from us as the planet pulls it — the radial-velocity method. Only when both measurements exist can you compute a density and say whether the planet is rock or gas.

When a planet crosses its star, the star dims by the ratio of the areas — nothing else. An Earth in front of the Sun blocks 84 parts per million, which is the number the Kepler mission was built around. Put in a planet, a star and a period and get the depth, the transit duration, the chance of the orbit being lined up at all, and an honest answer about whether a telescope on the ground could ever see it.

Last updated 11 Nov 2026

Transit depth, duration and detectability from the radius ratio

Depth = (planet radius / star radius) squared, and nothing to do with distance

Planet radius, in Earth radiiOne Earth radius is 6378.1 km (IAU 2015 nominal equatorial). Useful values: Neptune is 3.860 and Jupiter 11.209 Earth radii (NASA fact sheets, divided by the IAU nominal Earth radius). Above about 11 a real planet stops growing — adding mass to a gas giant compresses it rather than inflating it — so a radius of 20 here describes a star, not a planet.

Star radius, in solar radiiOne solar radius is 6.957 × 10⁸ m (IAU 2015 nominal). This matters as much as the planet does, and it is squared in the answer. A red dwarf of 0.2 solar radii gives the same planet twenty-five times the transit depth, which is exactly why small cool stars are where Earth-sized planets are actually found.

Orbital period, in daysThe period does not affect the depth at all — only the duration, the orbital distance and the odds of seeing a transit. The default is the sidereal year, 365.256 363 days. Hot Jupiters sit between 1 and 5 days, which is why they were the first planets found: a short period means a transit every few nights instead of once a year.

Star mass, in solar massesUsed only to turn the period into an orbital distance, through Kepler’s third law: a = (GM P²/4π²)^1/3. The third law page does that step on its own. A red dwarf is about 0.2 solar masses and 0.2 solar radii together, so changing one usually means changing both.

Impact parameter b, 0 for a central crossingHow far off-centre the planet crosses, measured in stellar radii: 0 is straight through the middle, 1 just clips the edge. Raise it and the chord gets shorter, so the transit gets briefer while the depth stays the same. Above b = 1 − k the planet is never fully on the disc, the transit is grazing and the real depth is less than this page reports; above b = 1 + k there is no transit at all.

Above, the star and the planet at true relative size, with the dashed line the chord the planet follows — raise b and watch the chord slide towards the edge and the transit shorten. Below, the light curve. The time axis is a fixed 36-hour window at ten pixels per hour, so the width of the dip is the real duration. The depth axis is logarithmic, because an Earth at 84 ppm and a hot Jupiter at 1% are a factor of 125 apart and will not share a linear picture; the real depth is printed beside it. The planet is never drawn smaller than about one pixel, which is itself the point: an Earth in front of the Sun is nine thousandths of the star’s width.

84.05ppmExample

The Earth crossing the Sun, as an alien astronomer would see it. Planet radius 1 Earth radius (6378.1 km), star radius 1 solar radius (6.957 × 10⁸ m), period 365.256 363 days, star mass 1 solar mass, and a central crossing (b = 0). How deep is the dip, how long does it last, and what are the odds of seeing it?

An area ratio, and a chord across a circle

δ = (R_p/R_star)² = k² · T_tot = (P/π) arcsin[(R_star/a)√((1+k)² − b²)] · P(transit) = R_star/a

δ: the fractional drop in brightness. It is the ratio of the two areas, πR_p² over πR_star², so the π cancels and the radii are squared. Distance does not appear anywhere.
k: R_p/R_star, the radius ratio. This is what a transit actually measures; the planet’s absolute size follows only once the star’s size is known from something else.
b: impact parameter: how far from the centre of the disc the chord runs, in units of the star’s radius. b = 0 is a central crossing and the longest transit.
a: orbital radius, worked out here from the period and the star’s mass by Kepler’s third law.
T_full: the flat-bottomed part, which uses (1 − k) in place of (1 + k). The difference between the two, halved, is the ingress time — how long the planet takes to get fully onto the disc.

Worked example

The Earth crossing the Sun, as an alien astronomer would see it. Planet radius 1 Earth radius (6378.1 km), star radius 1 solar radius (6.957 × 10⁸ m), period 365.256 363 days, star mass 1 solar mass, and a central crossing (b = 0). How deep is the dip, how long does it last, and what are the odds of seeing it?
The radius ratio, in the same units. k = R_p/R_star = 6.3781 × 10⁶ ÷ 6.957 × 10⁸ = 0.009 168. The Earth is about one part in 109 of the Sun’s width.
Square it, because the dip is an area ratio. δ = k² = 8.405 × 10⁻⁵ = 84.05 ppm, or 0.0084%. NASA’s Kepler mission design paper states that “an Earth-Sun analog transit produces a signal of 84 parts per million”, so the arithmetic is right. Notice what is not in the calculation: the distance to the star, its brightness, its temperature. Only the two radii.
Where the planet is, from the period. Kepler’s third law gives a = (GM P²/4π²)^1/3 = 1.496 × 10¹¹ m, which is 1 au as it must be. In units of the star’s own radius, a/R_star = 215. That number is worth keeping: the Sun is 215 solar radii away from us.
The odds of seeing the transit at all. For a randomly tilted orbit the probability is R_star/a = 1/215 = 0.465%. Fewer than one system in two hundred like ours is lined up well enough. That is why transit surveys watch hundreds of thousands of stars at once.
The duration. T_tot = (P/π) arcsin[(R_star/a)(1 + k)] with b = 0. The bracket is 0.004 650 × 1.009 17 = 0.004 693, whose arcsine in radians is 0.004 693 as well, because the angle is tiny. So T = (31 558 150/π) × 0.004 693 = 47 143 s = 13.10 hours. Taking the arcsine in degrees is the standard way to get this wrong, and it gives an answer 57 times too big.
And the ingress. Repeating with (1 − k) instead of (1 + k) gives the flat-bottomed part, 12.86 h. The difference, halved, is 7.1 minutes: that is how long the Earth takes to slide fully onto the Sun’s disc. Measuring that slope is how b is recovered from a real light curve.
So: one 13-hour dip of 84 ppm, once a year, with a 0.47% chance the geometry even allows it. That single sentence is the design specification for Kepler, and it is why the mission needed a telescope in space staring at the same patch of sky for years.

What each planet-and-star pairing actually gives you

Every row is this page, run at the radii in the first two columns. **Read the first two rows together.** The same Earth, in front of a star one fifth the size, gives twenty-five times the depth — because the depth goes as 1/R_star². That one line explains why almost every Earth-sized planet in the catalogues orbits a red dwarf, and why the TRAPPIST-1 system was found from the ground. The last row is the warning: a dip of 26% is not a planet, and telling planets from eclipsing binary stars is most of the work in a transit survey.
Planet	Star	k	Depth (ppm)	Where it can be seen
Earth (1 R⊕)	Sun (1 R☉)	0.0092	84	space only
Earth (1 R⊕)	red dwarf (0.2 R☉)	0.0458	2101	a good ground telescope
Neptune (3.860 R⊕)	Sun (1 R☉)	0.0354	1252	hard from the ground
Jupiter (11.209 R⊕)	Sun (1 R☉)	0.1028	10 560	an amateur telescope
Jupiter (11.209 R⊕)	red dwarf (0.2 R☉)	0.5138	264 000	not a planet transit at all — 26% is an eclipsing binary

The precision each kind of observation reaches

These four rows are the honest answer to “could I see this?”, and the gap between the second and third is the whole reason transit photometry moved into space. A wide-field survey camera on the ground is fighting the atmosphere, which scintillates by far more than a part in ten thousand; above the atmosphere the limit is just how many photons you collect. Figures for a single telescope watching one bright star fall between the extremes shown here, and this page cites no source for them, so none is quoted.
Instrument	Precision reached	Shallowest transit it can see	Source
Kepler, specification	20 ppm in 6.5 h at V = 12	about 20 ppm	Koch & Borucki et al. 2010, Table 1
Kepler, requirement it was built for	84 ppm detected reliably	an Earth round a Sun	the same paper, section 2.3
SuperWASP, wide-field survey from the ground	better than 1% for V brighter than 12.5	about 1% — 10 000 ppm	Christian, Pollacco et al. 2006
SuperWASP candidate selection	1 to 10 millimagnitudes	1000 to 10 000 ppm	the same paper, abstract

Why the depth is a ratio of areas, and what a transit cannot tell you

The whole method rests on one idea: a planet crossing its star blocks a fraction of the light equal to the ratio of the two discs. The star’s disc has area πR_star² and the planet’s silhouette has area πR_p², so the fraction blocked is (R_p/R_star)². The π cancels. The distance to the system cancels. The star’s brightness cancels. A transit measures a ratio, and that is why it is such a clean measurement — and also why it tells you the planet’s size only once somebody has measured the star’s size by other means.

The squaring is what makes small planets so hard. Jupiter is eleven times the Earth’s radius, so it blocks a hundred and twenty-five times as much light. A hot Jupiter gives a dip of about 1%, which an amateur telescope can record. An Earth gives 84 parts per million, which is a tenth of a millimagnitude, and the Earth’s own atmosphere makes stars twinkle by much more than that. There is no clever processing that recovers it from the ground; you have to leave.

The second thing a transit gives you is the duration, and the duration is a chord. The planet crosses a chord of the stellar disc, and how long that takes depends on how long the chord is — which depends on how far off-centre the crossing is — and on how fast the planet is moving. A central crossing is the longest. A grazing one is brief and has no flat bottom at all, because the planet is never fully in front. Fitting the shape of the light curve recovers both k and b, and that is how the diagram above is drawn.

Most planets are invisible to this method and always will be. A transit only happens if the orbit is nearly edge-on from where we sit, and the chance of that for a randomly tilted orbit is R_star/a — 0.47% for an Earth-like orbit and about 10% for a hot Jupiter three days out. So the method is biased, strongly and in a known direction: it over-reports big planets close to their stars, because those are both deeper and more likely to line up. Every statement about “how common” a kind of planet is has to correct for that before it means anything.

What a transit cannot tell you is the planet’s mass. The light curve knows nothing about how heavy the thing crossing is — a dense rock and a puffball of the same size give identical dips. Mass comes from the radial-velocity wobble of the star, a completely separate measurement. Put the two together and you get a density, and a density is what distinguishes a rocky world from a gas one. That is why the most valuable planets are the ones with both.

Where this goes wrong: radians. The arcsine in the duration formula returns radians, and every trigonometric function on this site works in radians, because that is the coherent SI unit of angle. A calculator left in degree mode gives a transit 57.3 times too long, and a 13-hour transit becomes a month, which at least looks wrong. The subtler error is mixing units in k: an Earth radius in kilometres divided by a solar radius in metres gives a ratio a thousand times too small and a depth a million times too small, and 0.000 08 ppm looks like a perfectly ordinary disappointing number. Put both radii into the same unit on a line of their own before dividing.

Frequently asked questions

Why does the transit depth not depend on how far away the star is?

Because it is a ratio. Moving the whole system further away makes the star fainter and the planet’s shadow smaller in exactly the same proportion, so the fraction of light blocked is unchanged. What distance does affect is how many photons you collect, and therefore how much noise sits on top of the dip — a faint star needs a longer stare for the same precision. The depth itself is a property of the two radii and nothing else.

How deep is the Earth crossing the Sun, and why does that number matter?

84 parts per million, which this page reproduces from the two radii. It matters because it was the design driver for the Kepler mission: NASA’s own mission design paper lists “reliably detect transits of 84 ppm in 6.5 hrs” as the third requirement, and specifies 20 ppm of photometric precision in 6.5 hours on a V = 12 Sun-like star to achieve it. Everything about the telescope follows from that one number.

Can I see an exoplanet transit from my back garden?

A hot Jupiter, yes, with a small telescope, a camera and patience: a 1% dip over two or three hours is within reach of careful amateur photometry, and amateurs contribute real timing measurements. Anything in the hundreds of parts per million is not, and anything like an Earth round a Sun is thousands of times below what the atmosphere allows. The band this page prints tells you which case you are in.

What is the impact parameter, and why does it shorten the transit?

It is how far from the centre of the stellar disc the planet’s path runs, measured in stellar radii. A central crossing, b = 0, follows the full diameter and takes the longest. Push b towards 1 and the chord becomes a short cut near the edge, so the transit gets briefer while the depth stays the same. Above b = 1 − k the planet never gets fully onto the disc, the flat bottom disappears, and the dip becomes a V — which is also how a grazing transit is recognised.

Does a transit tell me the planet’s mass?

No. It gives you the planet’s size relative to its star, and nothing about how heavy it is. A solid iron ball and a ball of fog of the same radius block exactly the same light. Mass comes from watching the star wobble towards and away from us as the planet pulls it — the radial-velocity method. Only when both measurements exist can you compute a density and say whether the planet is rock or gas.

Why are so many known planets large and close to their stars?

Because that is what the method can see, not because that is what is out there. Large planets give deeper dips (depth goes as the radius squared) and close-in planets transit far more often and are far more likely to be lined up at all (the probability is R_star/a). Both biases push the same way. Correcting for them is a standard and difficult part of every survey paper, and the raw catalogue should never be read as a census.

Related calculators

References

D. G. Koch, W. J. Borucki et al., Kepler Mission Design, Realized Photometric Performance, and Early Science, Astrophysical Journal Letters 713 (2010) L79, arXiv:1001.0268. States that “an Earth-Sun analog transit produces a signal of 84 parts per million (ppm)” and that the mission’s third requirement was “to reliably detect transits of 84 ppm in 6.5 hrs”, against a specification in its Table 1 of 20 ppm in 6.5 h for a V = 12 solar-like star over more than 90% of the field of view. This page’s default inputs reproduce the 84 ppm exactly, which is the check that its arithmetic is right.
J. N. Winn, Transits and Occultations, arXiv:1001.2010, a chapter in Exoplanets (ed. S. Seager, University of Arizona Press). Transit depth δ ≈ k² with k = R_p/R_star; total duration T_tot = (P/π) arcsin[(R_star/a)√((1+k)² − b²)/sin i]; the full (flat-bottomed) duration with (1−k) in place of (1+k); and the transit probability R_star/a.
D. J. Christian, D. L. Pollacco et al., The SuperWASP wide-field exoplanetary transit survey: candidates from fields 23hr < RA < 03hr, MNRAS 372 (2006) 1117, arXiv:astro-ph/0608142. A wide-field ground-based survey of about 400 000 stars at V = 8 to 13.5, reaching photometric precision “better than 1 per cent” for stars brighter than V = 12.5, and selecting its planet candidates from low-amplitude variability of about 1 to 10 millimagnitudes — which is 1000 to 10 000 ppm. That is the honest ground-based scale to compare this page’s depth against.
E. Mamajek et al., IAU 2015 Resolution B3 on Recommended Nominal Conversion Constants for Selected Solar and Planetary Properties, arXiv:1510.07674. Source of every GM and radius used here: nominal solar mass parameter 1.3271244 × 10²⁰ m³ s⁻², terrestrial 3.986004 × 10¹⁴, jovian 1.2668653 × 10¹⁷; nominal solar radius 6.957 × 10⁸ m, terrestrial equatorial radius 6.3781 × 10⁶ m, jovian equatorial 7.1492 × 10⁷ m. The resolution is explicit that these are conversion factors fixed by convention, not measurements with uncertainties.
BIPM, The International System of Units (SI), 9th edition (2019). The radian is the coherent SI unit of plane angle and has dimension one, which is why every trigonometric step on these pages is taken in radians; the standard atmosphere of 101 325 Pa exactly comes from Resolution 4 of the 10th CGPM (1954).