Answered By Albert S

Why doesn't the Earth crash into the Sun? Because of the Centipedal Force as it goes around it's orbit.

Why doesn't the Eath fly off into space? Because of the Gravitational force between the Sun and the Earth.

These two forces are balanced; equal magnitude -> opposite direction.

The Centipedal Corce is calculated as F_c=mv²/r

The Gravitational Force is calculated as F_g=GmM/r²

The only thing is we don't directly have the velocity. We can calculate it though. The Earth's orbit is almost circular so the velocity of the Earth can be calculated from the equation

d=vt or v=d/t

In one orbit around the Sun the distance travelled is 2 pi r (sorry I don't have a pi symbol since you are in a rush). The time needs to be in seconds so convert 365.25 days to seconds using 24 hours in a day, 60 minutes in an hour and 60 seconds in a minute. 

Substitute this new quantity (containing r) into the v in the Centrideal Force equation.

Equate F_c=F_g

Isolate r and then it is a a situation where you can use your calculator. (The mass of the Earth will cancel out one each side as well)

Answered By Vincent E

The earth is experiencing both gravitational force from the Sun, and centripetal force due to its orbit around the sun. These forces are equal. Therefore, Fc=Fg. We know that   $Fc=\frac{m_{Earth}v^2}{R}$Fc=m_Earthv²R  . From Newton's Law of Gravitation, we know that  $Fg=\frac{Gm_{Earth}m_{Sun}}{R^2}$Fg=Gm_Earthm_SunR² . If we make these equations equal, we get  $\frac{m_{Earth}v^2}{R}=\frac{Gm_{Earth}m_{Sun}}{R^2}$m_Earthv²R =Gm_Earthm_SunR² . Rearrange for R to get  $R=\frac{Gm_{Sun}}{v^2}$R=Gm_Sunv² . 

We also know that centripetal velocity is equal to  $v=\frac{2\pi R}{T}$v=2πRT‍ , where T is the orbital period in seconds and $2\pi r$2πr is the distance travelled in circular motion. We can substitute this into our R equation to get  $R=\frac{Gm_{sun}}{\left(\frac{2\pi R}{T}\right)^2}$R=Gm_sun(2πRT )² . Rearrange for R to get  $R=\left(\frac{Gm_{sun}T^2}{\left(2\pi\right)^2}\right)^{\frac{1}{3}}$R=(Gm_sunT²(2π)² )¹³ . At this point we can just plug and chug. Make sure to convert 365.25 days to seconds or else it won't work ( $365.25days\times\frac{24hours}{day}\times\frac{3600seconds}{hour}$365.25days×24hoursday ×3600secondshour ). Solve for  $R=1.496\times10^{11}m$R=1.496×10¹¹m . Which is essentially the same as the average Earth-Sun distance reported on your formula sheet.

Answered By Majid B

 $v=\frac{Circumference}{Period}=\frac{2\pi r}{T}$v=Circumƒ erencePeriod =2πrT  

 $F_c=F_g$F_c=F_g      =>      $\frac{mv^2}{r}=\frac{GmM}{r^2}$mv²r =GmMr²      =>        $r=\frac{GM}{v^2}=\frac{GM}{\left(\frac{2\pi r}{T}\right)^2}=\frac{GMT^2}{4\pi^2r^2}$r=GMv² =GM(2πrT )² =GMT²4π²r²        =>      $r^3=\frac{GMT^2}{4\pi^2}$r³=GMT²4π²      =>       $r=\sqrt[3]{\frac{GMT^2}{4\pi^2}}=\sqrt[3]{\frac{\left(6.67\times10^{-11}m^3.kg^{-1}.s^{-2}\right)\left(1.99\times10^{30}kg\right)\left(365.25\times24\times60\times60s\right)^2}{4\pi^2}}=1.496\times10^{11}m$r=³√GMT²4π² =³√(6.67×10^−‍11m³.kg⁻¹.s⁻²)(1.99×10³⁰kg)(365.25×24×60×60s)²4π² =1.496×10¹¹m 

    $F_c=F_g=\frac{GmM}{r^2}=\frac{\left(6.67\times10^{-11}m^3.kg^{-1}.s^{-2}\right)\left(5.97\times10^{24}kg\right)\left(1.99\times10^{30}kg\right)}{\left(1.496\times10^{11}m\right)^2}=3.541\times10^{22}N$F_c=F_g=GmMr² =(6.67×10⁻¹¹m³.kg⁻¹.s⁻²)(5.97×10²⁴kg)(1.99×10³⁰kg)(1.496×10¹¹m)² =3.541×10²²N