A planet of mass m = 1.55 x 1024 kg is orbiting in a circular path a star of mass M = 9.75 x 1029 kg. The radius of the orbit is R = 4.65 x 107 km. What is the orbital period (in earth days) of the planet? Where G = 6.67 � 10-11 N�m2/kg2 and 1 day = 8.54 � 104 s.
I used the formula
and got T = 247 s = 0.0289 days, but the online homework system
says thats the wrong answer.
What am I doing wrong?
T = 2*pi sqrt(r^3 / GM)
R=4.65*10^7km= 4.65*10^10metre
M=mass of star=9.75 x 1029 kg
T=7686902.828 sec or 90.01 days make sure u converted the radius in metres and took mass of star
T = 2*pi sqrt(r^3 / GM)
R=4.65*10^7km= 4.65*10^10metre
M=mass of star=9.75 x 1029 kg
G= 6.67
A planet of mass m = 1.55 x 1024 kg is orbiting in a circular path...
A planet of mass m = 6.25 × 1024 kg is orbiting in a circular path a star of mass M = 7.85 × 1029 kg. The radius of the orbit is R = 5.65 × 107 km. What is the orbital period (in Earth days) of the planet Tplanet?
A planet of mass m = 7.85 × 1024 kg is orbiting in a circular path a star of mass M = 3.85 × 1029 kg. The radius of the orbit is R = 8.35 × 107 km. What is the orbital period (in Earth days) of the planet Tplanet?
A planet of mass m=8.55×1024 kgm=8.55×1024 kg is orbiting in a circular path a star of mass M=9.35×1029 kgM=9.35×1029 kg . The radius of the orbit is R=3.75×107 kmR=3.75×107 km . What is the orbital period (in Earth days) of the planet TplanetTplanet ?
A planet of mass m=2.95×1024 kg is orbiting in a circular path a star of mass M=1.75×1029 kg . The radius of the orbit is R=6.25×107 km . What is the orbital period (in Earth days) of the planet Tplanet ?
A planet of mass 5.97 x 1024 kg revolves around a star of mass 2 x 1030 kg. Consider the orbit of the planet to be circular Ind the star is at the center. The distance between the star and the planet is 1.496 x 1011 m. Calculate the orbital velocity of the planet. (G= 6.67x10-11 m3 kg-152) (In the choices below, 10^4 is 104)
A planet of mass 5.97 x 1024 kg revolves around a star of mass 2 x 1030 kg. Consider the orbit of the planet to be circular and the star is at the center. The distance between the star and the planet is 1.496 x 1011 m. Calculate the orbital velocity of the planet. (G=6.67x10-11 mº kg1s 2) (In the choices below, 10^4 is 104) O 2.99 x 10^4 m/s 8.92 x 10^4 m/s O 4.58 x 10^4 m/s O...
A planet has a mass of 7.24 × 1024 kg and a radius of 7240 km. (a) Find the orbital radius of a satellite that is to complete 1 orbit every 3.4 hours. (b) Find the tangential velocity of a satellite that is to orbit 560 km above this planet’s surface. (c) Find the time period of a satellite that has an orbital radius of 9.35 × 107 m around this planet.
Planet X of mass mx = 6.3 × 1024 kg orbits S in uniform circular motion at a distance rx and with a period Px = 1.5 years (=4.7304 × 107 s). The mass of the star S is MS = 2 × 1031 kg and its radius is RS = 4.2 × 108 m, as shown in the figure. 1) What is rx, the orbital radius of Planet X around the Sun? a) rx = 4.21 × 109 m...
Question 4 1 pts Planet X has a mass of 5.97 x 1024 kg and it orbits 1.52 x 108 km from a star with a period of 412 days. What is the angular momentum of the planet around its star? Answer in units of kg m²/sec. Due to its large distance from its star, you can treat the planet as a point mass in its circular motion around the star.
A planet of mass 5 x 1024 kg is at location <5 x 1011, -4 x 1011, 0> m. A star of mass 9 x 1030 kg is at location <-6 x 1011, 9 1011, 0> m. It will be useful to draw a diagram of the situation, including the relevant vectors. (a) What is the relative position vector pointing from the planet to the star? (b) What is the distance between the planet and the star? (c) what is...