Wednesday

September 17, 2014

September 17, 2014

Posted by **Faith** on Thursday, January 13, 2011 at 1:09pm.

- Physics -
**tchrwill**, Friday, January 14, 2011 at 9:54amThe velocity required to keep the earth in a circulor orbit of radius 4 x(93,000,000) miles derives from

V = sqrt[µ/r] where µ = the Sun's gravitational constant = 4.68772x10^21 ft.^3/sec.^2 and r = the radius of the increased orbit in feet = 4.9104x10^11 feet.

Therefore, the orbital velocity becomes

V = sqrt[4.68772x10^21ft.^3/sec.^2/4.901x10^11] = 48,853 ft./sec. = mph. = 9.25 mph. compared to 97,706fps and 18.5 mph in its existing orbit

More symplictically, V = 97,700sqrt[r/4r] = 97,706[1/4) = 97,706/2 = 48,853 fps or 9.25 mph.

**Answer this Question**

**Related Questions**

physics - The mean distance of the planet Neptune from the Sun is 30.05 times ...

fizik asas - The earth travels in a nearly circular orbit about the sun. The ...

mit - A spacecraft of mass m is first brought into an orbit around the earth. ...

earth - planet is in circular orbit around the Sun. Its distance from the Sun ...

Physics - Use the following numbers for this question. MEarth 5.98e+24 kg ...

physics - the distance from the earth to the sun is 1.5 times 10 to the 11 power...

Physics - The earth moves around the sun in a nearly circular orbit of radius 1....

ALG/TRIG - Lost is an understatement..please help me understand this. Orbits and...

physics - The furthest distance from the Sun to Earth is df=1.521E8 The shortest...

physics - The earth moves around the sun in a nearly circular orbit of radius 1....