How Do Planetary Orbits Reveal the Masses of Stars and Heights of Satellites?

[coolfreesia]

1) In 2004 astronomers reported the discovery of a large Jupiter-sized planet orbiting very close to the star HD 179949 (hence the term "hot Jupiter"). The orbit was just 1/9 the distance of Mercury from our sun, and it takes the planet only 3.09 days to make one orbit (assumed to be circular). What is the mass of the star? Express your answer (a)in kilograms and (b)as a multiple of our sun's mass.
How fast (in km/s) is this planet moving?

2) The International Space Station makes 15.65 revolutions per day in its orbit around the earth. Assuming a circular orbit, how high is this satellite above the surface of the earth?

 

[foxjwill]

Please show some attempt at the solution. Once we see we're your trouble is, we'd be glad to help you.

 

[Moetasim]

1) To calculate the mass of the star, we can use the equation for orbital period: T = 2π√(a^3/GM), where T is the orbital period, a is the distance from the star, G is the gravitational constant, and M is the mass of the star. Plugging in the values given, we can solve for M: M = (4π^2a^3)/(GT^2). Assuming a circular orbit, we can use the distance of 1/9 the distance of Mercury from our sun, or approximately 0.000006 AU. Converting this to meters, we get 8.94 x 10^8 m. Plugging in the other values, we get M = 1.77 x 10^30 kg. As a multiple of our sun's mass, this would be approximately 0.89.

To calculate the speed of the planet, we can use the equation for orbital velocity: v = √(GM/a). Again, assuming a circular orbit, we can use the distance of 0.000006 AU, or 8.94 x 10^8 m. Plugging in the other values, we get v = 1.02 x 10^5 m/s, or approximately 102 km/s.

2) To calculate the height of the International Space Station (ISS), we can use the equation for orbital period: T = 2π√(r^3/GM), where T is the orbital period, r is the distance from the center of the Earth, G is the gravitational constant, and M is the mass of the Earth. We know that the ISS makes 15.65 revolutions per day, or one revolution every 92 minutes. Converting this to seconds, we get T = 5520 seconds. Plugging in the other values, we get r = 4.28 x 10^6 m. Subtracting the radius of the Earth (6.37 x 10^6 m), we get the height of the ISS above the surface of the Earth to be approximately 3.59 x 10^6 m, or 3590 km.