Question

Given that the mass of the sun and the earth are respectively 1.99x10^30 kg and 5.98x10^24...

Given that the mass of the sun and the earth are respectively 1.99x10^30 kg and 5.98x10^24 kg, and that, the distance between them is 1.50x10^11 m, (a) find the graviational attraction between the sun and the earth. (b) Taking one year to equal 3.156x10^7 s, and that the orbit of the earth around the sun is circular, find, in miles per hour, the speed (considered uniform) of the earth around the sun.

Homework Answers

Answer #1

(a) We know that the force due to gravitational attraction is given by
F = GMm/r2
where G is gravitational constant = 6.67*10-11
M is mass of sun = 1.99*1030 kg
m is mass of earth = 5.98*1024 kg
r is distance between them = 1.5*1011 m
F = (6.67*10-11 )*( 1.99*1030 *5.98*1024)/(1.5*1011 )2 = 3.528*1022 N
(b) We know that
Linear speed is given by (V)= 2r/T
Where T is the time period to complete one revolution = 3.156*107 s
V = (2*1.5*1011 )/(3.156*107 ) = 29863.05 m/s
Now in miles /hour
1 m/s = 2.23694 mi/hr
V =  29863.05*2.23694 = 66801.74 mi/hr

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Earth has a mass of 5.97 * 10^24 kg and a radius of 6.38 *10^6 m....
Earth has a mass of 5.97 * 10^24 kg and a radius of 6.38 *10^6 m. Assume it is a uniform solid sphere. The distance of Earth from the Sun is 1.50 * 10^11 m. (Assume Earth completes a single rotation in 24.0 hours and orbits the Sun once every 365 Earth days.) (a) Calculate the angular momentum of Earth in its orbit around the Sun. (b) Calculate the angular momentum of Earth on its axis. Please show your work.
A satellite of mass 1525 kg is in circular orbit around Earth. The radius of the...
A satellite of mass 1525 kg is in circular orbit around Earth. The radius of the orbit of the satellite is equal to 1.5 times the radius of Earth (RE = 6.378*106 m, ME = 5.98*1024 kg, G = 6.67*10-11 Nm2/kg2). (a) Find the orbital period of the satellite? (b) Find the orbital (tangential) velocity of the satellite.  (c) Find the total energy of the satellite?
(a) Using elementary Newtonian mechanics find the period of a mass m 1in a circular orbit...
(a) Using elementary Newtonian mechanics find the period of a mass m 1in a circular orbit of radius r around a fixed mass m 2. (b) Using the separation into CM and relative motions, find the corresponding period for the case that m 2is not fixed and the masses circle each other a constant distance r apart. Discuss the limit of this result if m 2oo. (c) What would be the orbital period if the earth were replaced by a...
A satellite of mass 350 kg is in a circular orbit around the Earth at an...
A satellite of mass 350 kg is in a circular orbit around the Earth at an altitude equal to the Earth's mean radius. (a) Find the satellite's orbital speed. m/s (b) What is the period of its revolution? min (c) Calculate the gravitational force acting on it. N
A planet is in a circular orbit around the sun. Use Newton's law of gravity and...
A planet is in a circular orbit around the sun. Use Newton's law of gravity and his second law of motion to calculate the period of the planet (in day). Data: Mass of sun = 1.989 e+30 kg; Mass of planet = 6.0 e+24 kg; Orbit radius = 1.496 e+11 m.
A comet of mass 1.20 ✕ 1010 kg moves in an elliptical orbit around the Sun....
A comet of mass 1.20 ✕ 1010 kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.200 AU and 62.0 AU. (Note: 1 AU = one astronomical unit = the average distance from the Sun to Earth = 1.496 ✕ 1011 m.) (a) What is the eccentricity of its orbit? (b) What is its period? ______yr (c) At aphelion what is the potential energy of the comet-Sun system? _____J
Mass of earth: 5.972x10^24 kg Mass of Mangalyaan: 1,337.2 kg Radius of earth: 6,378 km Apogee:...
Mass of earth: 5.972x10^24 kg Mass of Mangalyaan: 1,337.2 kg Radius of earth: 6,378 km Apogee: The point in the orbit of the moon or satellite at which it is furthest from the earth. On Nov. 5th of 2013, a space probe known as Mangalyaan was launched and it reached its apogee of 23,500 km. Mangalyaan, then went through 7 different apogee raising maneuvers before its trans-mars injection on Nov. 30th, 2013. Before its final maneuver to trans-mars injection, the...
Mass of earth: 5.972x10^24 kg Mass of Mangalyaan: 1,337.2 kg Radius of earth: 6,378 km Apogee:...
Mass of earth: 5.972x10^24 kg Mass of Mangalyaan: 1,337.2 kg Radius of earth: 6,378 km Apogee: The point in the orbit of the moon or satellite at which it is furthest from the earth. On Nov. 5th of 2013, a space probe known as Mangalyaan was launched and it reached its apogee of 23,500 km. Mangalyaan, then went through 7 different apogee raising maneuvers before its trans-mars injection on Nov. 30th, 2013. Before its final maneuver to trans-mars injection, the...
assume that the earth orbits the sun in a circular orbit with constant speed at a...
assume that the earth orbits the sun in a circular orbit with constant speed at a radius of 1.5 x 10^11 m with a period of T = 356 days. Calculate the speed of the earth in its orbit (in m/s)
A satellite is in a circular orbit around the Earth at an altitude of 3.32 106...
A satellite is in a circular orbit around the Earth at an altitude of 3.32 106 m. (a) Find the period of the orbit. (Hint: Modify Kepler's third law so it is suitable for objects orbiting the Earth rather than the Sun. The radius of the Earth is 6.38 106 m, and the mass of the Earth is 5.98 1024 kg.) h (b) Find the speed of the satellite. km/s (c) Find the acceleration of the satellite. m/s2 toward the...