Calculus Problem in Astronomy Class. Need someone to check my solution. ?

Question

Assume that the earth moves around the sun in a circular orbit with a 92 million mile radius. Assume further that it takes 365 days to make one full circle. Then its orbital velocity, neglecting all other celestial motions, such as the motion of the solar system toward Vega or galactic rotation, etc., would be, in miles per second, correct to the nearest whole mile per second,


..... 27 miles per second ?

Answer 1

v ≈ 2πr / T

r = distance
T = orbital period

v ≈ (2)(π)(92,000,000) / T

T = (365 days) (24hours/day) (60mins/hour) (60 secs/min)
T = 31536000 secs

v ≈ (2)(π)(92,000,000 miles) / (31536000 sec)

≈ 18.33 mi/sec

Answer 2

The total length of the orbit would be the perimeter of a circle of radius 92 million miles. The length is 2πr = 2π(92000000) = 184000000π = about 578053048.26 miles.

Since you're looking for a speed in miles per second, you need to know how many seconds are in 365 days. This is 365(24)(60)(60) = 31536000 seconds.

As to the speed, 578053048.26 / 31536000 = about 18.33 miles per second. To the nearest whole unit, it's 18 miles per second.

Answer 3

This is no calculus problem, it's algebra.

Radius = 92 x 10^6 miles

Circumference = 3.14 x 2 x (92 x 10^6)= 5.776 x 10^8 miles

Speed = Circumference/(365.25*24*60*60)

Speed = 18.31 miles per second

Answer 4

Angular velocity = w = (2π / 365) radians / day
r = 92 x 10^6 miles
v = w r
v = (2π / 365) x 92 x 10^6 miles / day
v = (2π/365) x 92 x 10^6 / (365 x 24 x 3600) miles / sec
v = 18.3 miles/sec (to 1 dec. place)

Answer 5

(3.14x(92,000,000x2))
/(365*24*60*60)

3.14= pi
92,000,000 = Radius
92,000,000*2 = Diameter
365 days
24 hours
60 minutes
60 seconds

=18.3206494 miles/second

You have to get the circumference of the circle in order to do this.

Answer 6

perimeter of orbit
92*10^6*2*pi mille radiusin a year

a year =365*24*3600 seconds

so in 1 second earth moves
92*10^6*2*pi/365*24*3600=18.32miles pes second

i hope my calculus is good

Answer 7

you have a nil over 0, so which you may desire to locate a topic-loose element interior the ideal and backside, which happens to be x+3. you may desire to do man made branch for the backside equation to locate the different element of the denominator. you may get x^2-3x+9. You bypass out the x+3 on the ideal and the backside of the equation after which you plug a -3 into the equation. you may get a -a million/4.