## Alberta Free Tutoring And Homework Help For Physics 20

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### The Hubble Spcae telesope orbits the Earth every 95 mins. Calc the altitude abover earths surface.

3 years ago The Hubble Space Telescope is experiencing both gravitational force from the Earth, and centripetal force due to its orbit around the Earth. These forces have to be equal, otherwise the telescope would not stay in orbit. Therefore, Fc=Fg. We know that    $Fc=\frac{m_{Hubble}v^2}{R}$Fc=mHubblev2R   . From Newton's Law of Gravitation, we know that   $Fg=\frac{Gm_{Hubble}m_{Earth}}{R^2}$Fg=GmHubblemEarthR2  . If we make these equations equal, we get   $\frac{m_{Hubble}v^2}{R}=\frac{Gm_{Hubble}m_{Earth}}{R^2}$mHubblev2R =GmHubblemEarthR2 . Rearrange for R to get   $R=\frac{Gm_{Earth}}{v^2}$R=GmEarthv2

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   $R=\frac{Gm_{Earth}}{\left(\frac{2\pi R}{T}\right)^2}$R=GmEarth(2πRT )2  . Rearrange for R to get   $R=\left(\frac{Gm^{_{_{Earth}}}T^2}{\left(2\pi\right)^2}\right)^{\frac{1}{3}}$R=(GmEarthT2(2π)2 )13  . At this point we can just plug and chug. Make sure to convert 95 mins to seconds or else it won't work ($95mins\times\frac{60sec}{min}$95mins×60secmin ). Solve for   $R=6894403.754m$R=6894403.754m. Note that R is the orbital radius of the satellite, meaning the distance from the satellite to the center of the Earth. To find the altitude, which is the distance of the satellite from the Earth's surface, we have to subtract the radius of the Earth from our calculated R. Therefore,  $Altitude=6894403.754m-6.38\times10^6m=514403.7541m$Altitude=6894403.754m6.38×106m=514403.7541m. In kilometres, it would be 514.4 km. Considering significant digits (2 from the question) you could also write it as  $5.1\times10^2$5.1×102 km.

3 years ago $F_c=F_g$Fc=Fg   =>   $\frac{mv^2}{r}=\frac{GmM}{r^2}$mv2r =GmMr2 .
$v=\frac{d}{t}=\frac{\text{Circumference}}{Period}=\frac{2\pi r}{T}$v=dt =CircumferencePeriod =2πrT    =>   $\frac{m\left(\frac{4\pi^2r^2}{T^2}\right)}{r}=\frac{GmM}{r^2}$m(4π2r2T2 )r =GmMr2    =>   $r^3=\frac{GMT^2}{4\pi^2}$r3=GMT24π2    =>
$r=\sqrt{\frac{GMT^2}{4\pi^2}}==\sqrt{\frac{\left(6.67\times10^{-11}N.m^2.kg^{-2}\right)\left(5.97\times10^{24}kg\right)\left(95\times60s\right)^2}{\left(4\pi^2\right)}}=6894404m$r=3GMT24π2 ==3(6.67×1011N.m2.kg2)(5.97×1024kg)(95×60s)2(4π2) =6894404m
$r=Radius+Altitude=R+A$r=Radius+Altitude=R+A  =>    $A=r-R=6894404m-6.37\times10^6m=524404m\approx524km$A=rR=6894404m6.37×106m=524404m524km