Question

Problem 9: (Exoplanets) The HAT-P-7 stars are stars on the F sequence. From the latest findings, found that A planet (exoplanet) orbits HAT-P-7 Information on the stars HAT-P-7 The apparent magnitude m = 10.46 Mass M = 1.53 Mo Radius R = 1.98 RO Parallax angle p = 0.0029" The two graphs below show a light curve when the planets block the stars (eclipsing) 1.0021 1.000 1.002 1.000 0.998 0.998 Relative Brightness Relative Brightness 0.996 996 0.994 0.992! 0.9941 -3 -2 No Hours from Midtransit 0.992. Time days) (A) Find the orbital period of this planet (B) Calculate the orbital radius of this planet (C) Calculate the radius of this planet

Answer 1

Solution of (A):

The time between two successive transits is planet's orbital period which is the length of that planet's year:

1002 1.000 0998 Relative Brightness 0.996 Orbital Period T 0.994 0.992 2 3 Time days)

Therefore, the orbital period :

P= 3.4 - 1.2

P= 2.2 days

Solution of (B):

The planet's orbital radius is given by:

1/3 a = PGM, 472

Where,
Period of the planet
→ P= 2.2 days 2.2 days = 190080 sec

Gravitational constant
→G= 6.673 x 10-11 N m2 kg

Mass of Star
→ M, M. = 1.53MO

Mass of Sun
→ MO = 1.9891 x 1030 kg

1/3 ..a = PGM, 472

1.a (190080)2 x 6.673 x 10-11 x 1.53 x 1.9891 x 1030 1/3 472

:.a = 57.06818945 x 108 m

1.a = 57.1 x 10 km

Solution of (C):

Here,
Radius of the star
+R, = 1.98RO

Radius of Sun
RO = 696340 x 100 m

Radius of the planet
→ Rp

Since, the depth planet's transits reveal the size of the planet, the depth in the brightness is observed from the graph is:

1002 1.000 0998 Relative Brightness Depth D 0.996 0.994 0.992 2 3 Time days)

D=1 - 0.9935

D=0.0065

The brightness depth is given by

D = Rp R

:.R=RD

:. R, = RVD

..Rp = 1.98 x 696340 x 103 x 70.0065

:: Rp = 11.11586367 x 10 m

:: Rp = 11.116 x 10 m

Which
R
is approximately equal to 1.5 times Jupiter's radius.