Question

1) An exoplanet orbits a star in an ellipse with semi-major axis 0.81⨯10¹² m and an eccentricity of 0.53.
What is the distance from the star to the opposite focus of the ellipse?

2)A satellite is placed in geosynchronous orbit around an exoplanet of mass 67⨯10²⁴ kg and radius 73⨯10⁶ m.
If the planet's day is 95.6 hours, at what altitude should you place the satellite?

Answer 1

distance from the star to the opposite focus of the ellipse = 2*e*a

e = eccentricity = 0.53

a = lenght of semi major axis = 0.81*10^12 m

distance d = 2*0.53*0.81*10^12 = 85.86*10^10 m

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2)

time period of a planet T = 2*pi*(R+h)^(3/2)/sqrt(G*M)

R = radius of exoplanet = 73*10^6 m

M = mass of exoplanet = 67*10^24 kg

time petiod T = 95.6 hours = 95.6*3600 s

95.6*3600 = 2*pi*(73*10^6 + h)^(3/2)/sqrt(6.67*10^-11*67*10^24)

altitude h = 164.5*10^6 m <<<<---------ANSWER