The Transit Method (Kepler Space Telescope) is used to find out the size. The dip in brightness of a star by a planet is usually very less unlike a binary system.

• F₀ is flux of the star before the planet occults it.

• F₁ is the flux after the entire planet is in front of star.

The following image will be used for all the calculations.

$$frac{F_0 – F_1}{F_0} = frac{pi r_p^{2}}{pi R^2_ast}$$

$$frac{Delta F}{F} cong frac{r^2_p}{R^2_ast}$$

$$left ( frac{Delta F}{F} right )_{earth} cong 0.001%$$

$$left ( frac{Delta F}{F} right )_{jupiter} cong 1%$$

This is not easy to achieve by ground based telescope. It is achieved by the Hubble telescope.

Here, $t_T$ is time between position A and D and $t_F$ is time between position B and C.

The geometry of a transit related to the inclination i of the system. Transit latitude and inclination are interchangeable.

From the above images, we can write −

$$frac{h}{a} = cos(i)$$

$$frac{h}{R_ast} = sin(delta)$$

$$cos(i) = frac{R_ast sin(delta)}{a}$$

$$y^2 = (R_ast + R_p)^2 – h^2$$

$$y = [(R_ast + R_p)^2 – h^2]^{frac{1}{2}}$$

$$sin(theta) = frac{y}{a}$$

$$theta = sin^{-1}left [ frac{(R_ast + R_p)^2 – a^2cos^2(i)}{a^2} right ]^{frac{1}{2}}$$

$$t_T = frac{P}{2pi} times 2theta$$

Here, $t_T$ is the fraction of a time-period for which transit happens and (2θ/2π) is fraction of angle for which the transit happens.

$$sin(frac{t_Tpi}{P}) = frac{R_ast}{a}left [ left ( 1+ frac{R_p}{R_ast}right )^2 – left ( frac{a}{R_ast}cos(i)right )^2 right ]^{frac{1}{2}}$$

Usually, a >> R∗ >> Rp. So, we can write −

$$sin(frac{t_Tpi}{P}) = frac{R_ast}{a}left [ 1- left ( frac{a}{R_ast}cos(i) right )^2right ]^{frac{1}{2}}$$

Here, P is the duration between two successive transits. The transit time is very less compared to the orbital time-period. Hence,

$$t_T = frac{P}{pi}left [ left ( frac{R_ast}{a}right )^2 – cos^2(i)right ]^{frac{1}{2}}$$

Here, t_T, P, R∗ are the observables, a and i should be found out.

Now,

$$sin(frac{t_Fpi}{P}) = frac{R_ast}{a}left [left (1 – frac{R_p}{R_ast} right )^2 – left ( frac{a}{R_ast}cos:i right )^2right ]^{frac{1}{2}}$$

where, $y^2 = (R_ast − R_p)^2 − h^2$.

Let,

$$frac{Delta F}{F} = D = left ( frac{R_p}{R_ast} right )^2$$

Now, we can express,

$$frac{a}{R_ast} = frac{2P}{pi} D^{frac{1}{4}}(t^2_T – t^2_F)^{-frac{1}{2}}$$

For the main sequence stars,

$$R_ast propto M^alpha_ast$$

$$frac{R_ast}{R_0} propto left ( frac{M_ast}{M_0}right )^alpha$$

This gives R∗.

Hence, we get value of ‘a’ also.

So, we get ‘R_p’, ’ap’ and even ‘i’.

For all this,

$$h leq R_ast + R_p$$

$$a: cos: i leq R_ast + R_p$$

For even 𝑖~89 degrees, the transit duration is very small. The planet must be very close to get a sufficient transit time. This gives a tight constraint on ‘i’. Once we get ‘i’, we can derive ‘m_p′from the radial velocity measurement.

This detection by the transit method is called as chance detection, i.e., probability of observing a transit. Transit probability (probability of observing) calculations are shown below.

The transit probability is related to the solid angle traced out by the two extreme transit configurations, which is −

$$Solid : angle :of :planet : = 2pi left ( frac{2R_ast}{a} right )$$

As well as the total solid angle at a semi-major axis a, or −

$$Solid :angle :of :sphere : =: 4pi$$

The probability is the ratio of these two areas −

$$= : frac{area :of: sky : covered :by: favourable : orientation}{area: of: sky :covered: by: all: possible: orientation: of: orbit}$$

$= frac{4pi a_pR_ast}{4pi a^2_p} = frac{R_ast}{a_p}$ $frac{area: of: hollow : cyclinder}{area: of: sphere}$

This probability is independent of the observer.

Points to Remember

• The Transit Method (Kepler Space Telescope) is used to find out the size.
• Detection by Transit Method is a chance detection.
• The planet must be very close to get sufficient transit time.
• Transit probability is related to the solid angle of planet.
• This probability is independent of observer frame of reference.