Kepler's 3rd law: ratios don't fit data












2












$begingroup$


I have been looking at satellite orbits around the earth, or any object around any planet in fact, and am trying to find the orbital radius, or semi major length of a given satellite.



Kepler's third law gives the equation $P^2 = a^3$ where $P$ is the period of orbit and $a$ the distance.



I have a table of satellites currently orbiting the earth, as well as their altitude in the sky on their geosynchronous trajectory. One in particular is 99.9 and has an altitude of 705.



By solving the equation for $a$, I get $a = (P^2)^{1/3}$.



When I plug in the numbers, they don't correspond.



So my questions are:




  1. Are there unit standards I need for both $P$ and $a$? Currently $P$ is in minutes, $a$ in kilometres.

  2. Am I missing something, like Newton's universal gravitational constant? I get a page deriving Kepler's third law using this constant.






newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites







share|cite|improve this question









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triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
    $endgroup$
    – Kyle Kanos
    7 hours ago










  • $begingroup$
    The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
    $endgroup$
    – jacob1729
    4 hours ago






  • 1




    $begingroup$
    BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
    $endgroup$
    – G. Smith
    1 hour ago


















2












$begingroup$


I have been looking at satellite orbits around the earth, or any object around any planet in fact, and am trying to find the orbital radius, or semi major length of a given satellite.



Kepler's third law gives the equation $P^2 = a^3$ where $P$ is the period of orbit and $a$ the distance.



I have a table of satellites currently orbiting the earth, as well as their altitude in the sky on their geosynchronous trajectory. One in particular is 99.9 and has an altitude of 705.



By solving the equation for $a$, I get $a = (P^2)^{1/3}$.



When I plug in the numbers, they don't correspond.



So my questions are:




  1. Are there unit standards I need for both $P$ and $a$? Currently $P$ is in minutes, $a$ in kilometres.

  2. Am I missing something, like Newton's universal gravitational constant? I get a page deriving Kepler's third law using this constant.






newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites







share|cite|improve this question









New contributor




triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
    $endgroup$
    – Kyle Kanos
    7 hours ago










  • $begingroup$
    The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
    $endgroup$
    – jacob1729
    4 hours ago






  • 1




    $begingroup$
    BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
    $endgroup$
    – G. Smith
    1 hour ago
















2












2








2





$begingroup$


I have been looking at satellite orbits around the earth, or any object around any planet in fact, and am trying to find the orbital radius, or semi major length of a given satellite.



Kepler's third law gives the equation $P^2 = a^3$ where $P$ is the period of orbit and $a$ the distance.



I have a table of satellites currently orbiting the earth, as well as their altitude in the sky on their geosynchronous trajectory. One in particular is 99.9 and has an altitude of 705.



By solving the equation for $a$, I get $a = (P^2)^{1/3}$.



When I plug in the numbers, they don't correspond.



So my questions are:




  1. Are there unit standards I need for both $P$ and $a$? Currently $P$ is in minutes, $a$ in kilometres.

  2. Am I missing something, like Newton's universal gravitational constant? I get a page deriving Kepler's third law using this constant.






newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites







share|cite|improve this question









New contributor




triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I have been looking at satellite orbits around the earth, or any object around any planet in fact, and am trying to find the orbital radius, or semi major length of a given satellite.



Kepler's third law gives the equation $P^2 = a^3$ where $P$ is the period of orbit and $a$ the distance.



I have a table of satellites currently orbiting the earth, as well as their altitude in the sky on their geosynchronous trajectory. One in particular is 99.9 and has an altitude of 705.



By solving the equation for $a$, I get $a = (P^2)^{1/3}$.



When I plug in the numbers, they don't correspond.



So my questions are:




  1. Are there unit standards I need for both $P$ and $a$? Currently $P$ is in minutes, $a$ in kilometres.

  2. Am I missing something, like Newton's universal gravitational constant? I get a page deriving Kepler's third law using this constant.






newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites




newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites






share|cite|improve this question









New contributor




triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









Qmechanic♦

108k122001253




108k122001253






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Check out our Code of Conduct.









asked 7 hours ago









triple7triple7

133




133




New contributor




triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
    $endgroup$
    – Kyle Kanos
    7 hours ago










  • $begingroup$
    The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
    $endgroup$
    – jacob1729
    4 hours ago






  • 1




    $begingroup$
    BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
    $endgroup$
    – G. Smith
    1 hour ago




















  • $begingroup$
    hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
    $endgroup$
    – Kyle Kanos
    7 hours ago










  • $begingroup$
    The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
    $endgroup$
    – jacob1729
    4 hours ago






  • 1




    $begingroup$
    BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
    $endgroup$
    – G. Smith
    1 hour ago


















$begingroup$
hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
$endgroup$
– Kyle Kanos
7 hours ago




$begingroup$
hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
$endgroup$
– Kyle Kanos
7 hours ago












$begingroup$
The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
$endgroup$
– jacob1729
4 hours ago




$begingroup$
The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
$endgroup$
– jacob1729
4 hours ago




1




1




$begingroup$
BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
$endgroup$
– G. Smith
1 hour ago






$begingroup$
BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
$endgroup$
– G. Smith
1 hour ago












3 Answers
3






active

oldest

votes


















4












$begingroup$

that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.



Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)






share|cite|improve this answer











$endgroup$









  • 5




    $begingroup$
    Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
    $endgroup$
    – triple7
    7 hours ago










  • $begingroup$
    @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
    $endgroup$
    – Jerry Schirmer
    6 hours ago






  • 1




    $begingroup$
    Yep, finally found it. Thanks
    $endgroup$
    – triple7
    6 hours ago



















1












$begingroup$

The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.



Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.



Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.






    share|cite|improve this answer









    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.



      Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)






      share|cite|improve this answer











      $endgroup$









      • 5




        $begingroup$
        Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
        $endgroup$
        – triple7
        7 hours ago










      • $begingroup$
        @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
        $endgroup$
        – Jerry Schirmer
        6 hours ago






      • 1




        $begingroup$
        Yep, finally found it. Thanks
        $endgroup$
        – triple7
        6 hours ago
















      4












      $begingroup$

      that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.



      Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)






      share|cite|improve this answer











      $endgroup$









      • 5




        $begingroup$
        Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
        $endgroup$
        – triple7
        7 hours ago










      • $begingroup$
        @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
        $endgroup$
        – Jerry Schirmer
        6 hours ago






      • 1




        $begingroup$
        Yep, finally found it. Thanks
        $endgroup$
        – triple7
        6 hours ago














      4












      4








      4





      $begingroup$

      that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.



      Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)






      share|cite|improve this answer











      $endgroup$



      that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.



      Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 6 hours ago

























      answered 7 hours ago









      Jerry SchirmerJerry Schirmer

      31.7k257107




      31.7k257107








      • 5




        $begingroup$
        Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
        $endgroup$
        – triple7
        7 hours ago










      • $begingroup$
        @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
        $endgroup$
        – Jerry Schirmer
        6 hours ago






      • 1




        $begingroup$
        Yep, finally found it. Thanks
        $endgroup$
        – triple7
        6 hours ago














      • 5




        $begingroup$
        Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
        $endgroup$
        – triple7
        7 hours ago










      • $begingroup$
        @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
        $endgroup$
        – Jerry Schirmer
        6 hours ago






      • 1




        $begingroup$
        Yep, finally found it. Thanks
        $endgroup$
        – triple7
        6 hours ago








      5




      5




      $begingroup$
      Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
      $endgroup$
      – triple7
      7 hours ago




      $begingroup$
      Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
      $endgroup$
      – triple7
      7 hours ago












      $begingroup$
      @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
      $endgroup$
      – Jerry Schirmer
      6 hours ago




      $begingroup$
      @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
      $endgroup$
      – Jerry Schirmer
      6 hours ago




      1




      1




      $begingroup$
      Yep, finally found it. Thanks
      $endgroup$
      – triple7
      6 hours ago




      $begingroup$
      Yep, finally found it. Thanks
      $endgroup$
      – triple7
      6 hours ago











      1












      $begingroup$

      The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.



      Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.



      Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.



        Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.



        Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.



          Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.



          Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!






          share|cite|improve this answer











          $endgroup$



          The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.



          Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.



          Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 6 hours ago

























          answered 7 hours ago









          swickrotationswickrotation

          615




          615























              0












              $begingroup$

              Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.






                  share|cite|improve this answer









                  $endgroup$



                  Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 7 hours ago









                  my2ctsmy2cts

                  5,9492719




                  5,9492719






















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                      GameSpot

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                      body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} GameSpot 戰地風雲：惡名昭彰2在Gamespot的評論 网站类型 新聞 持有者 CBS 创始人 Pete Deemer Vince Broady Jon Epstein 网站 http://www.gamespot.com/ 注册 Optional (free and paid) 推出时间 1996年5月1日 [1] GameSpot （中国大陆：游戏基地），於1996年5月由Pete Deemer和Vince Broady創立，是一個提供新聞、評論、預告片、下載及其他的相關資訊的電子遊戲網站。GameSpot被一間後來被CNET收購的企業ZDNet所收購。根據Alexa，GameSpot.com是200個網路擁擠最嚴重的網站之一。 除了由GameSpot員工創作的內容，網站還允許用戶寫評論、網誌、之後在網路論壇分享。一些在CNET旗下的GameFAQs分享。 2004年， GameSpot被Spike TV的觀眾選上「電子遊戲賞節目」贏得「最傑出遊戲網站。 [2] 其他的遊戲網站還有IGN、1UP.com、GameSpy是它最大的競爭對手。2008年，根據Compete.com的統計，「gamespot.com」吸引了最少6000萬人的點擊率。 [3] GameSpot的主頁鏈結了到最近新聞、評論、預告、和一些有關遊戲機的入口：Wii、任天堂DS、電腦遊戲、Xbox 360、PSP、PlayStation 2、PlayStation 3。它還有一列「最受歡迎遊戲名單」，還有給用戶快速獲得遊戲資訊的搜尋器。GameSpot 還包括一些小範圍的遊戲機：任天堂64、GameCube、Game Boy Color、Game Boy Advance、Xbox、PlayStation、SEGA Saturn、Dreamcast、Neo Geo Pocket Color、N-Gage、手機遊戲。 目录 1 歷史 1.1 國際歷史 1.2 著名的員工 2 評論和分...
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                      日野市

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                      body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 日野市 日野市 日文轉寫  • 日文 日野市  • 平假名 ひのし  • 罗马字 Hino-shi 金剛寺（關東三十六不動尊靈場第9號）的不動堂 市旗 徽章 日野市在東京都的位置 日野市 日野市在日本的位置 坐标： 35°40′17″N 139°23′42″E  /  35.6714°N 139.395°E  / 35.6714; 139.395 国家   日本 地方 關東地方 都道府縣 東京都 接鄰行政區 府中市、國立市、立川市、昭島市、八王子市、多摩市 政府  • 市長 大坪冬彦 面积  •  总计 27.53 平方公里（10.63 平方英里） 人口 （2014年12月1日）  • 總計 183,323  • 密度 6,660/平方公里（17,200/平方英里） 象徵  • 市樹 樫  • 市花 菊花  • 市鳥 普通翠鳥 时区 日本標準時間 （UTC+9） 地方公共團體編號 13212-8 邮政编码 〒 191-8686 市役所地址 日野市神明一丁目12番1號 電話號碼 +81-42-585-1111 法人編號 1000020132128 網站 http://www.city.hino.lg.jp/ 人口：日野市官方網頁 日野市 （日语： 日野市 ／ ひのし   Hino shi   * / ? ）為一位于東京都（不含島嶼部分）中央地帶的城市。從東京站乘坐中央線特別快速列車45分鐘即可到該市。面積27.53km²。1963年（昭和38年）11月3日，該市開始實行市制，為全日本第559個市。汽車企業日野自動車的總部位於這裡。 往東京都特別區部的通勤率為20.9%，往八王子市的通勤率為12.0%（平成22年國勢調査）。 目录 1 概要 ...
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                      Tu-95轟炸機

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                      body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 图-95 Ту–95 北约代号：熊（Bear） 一架在安格斯空军基地的图-95MS 概觀 類型 战略轰炸机 代號 北约代号： Bear （熊） 乘員 7名 駕駛員2名，機尾炮手1名，其他人員4名 首飛 1952年11月12日 服役 1956年 設計 圖波列夫設計局 產量 超过500架 現況 现役 主要用戶   蘇聯   俄羅斯 衍生機型 图-114、图-119、Tu-95 技术数据 長度 49.50米（162呎5吋） 翼展 51.10米（167呎8吋） 高度 12.12米（39呎9吋） 翼面積 310平方米（3,330平方呎） 空重 90,000公斤（198,000磅） 最大起飛重量 188,000公斤（414,500磅） 發動機 4具库兹涅佐夫NK-12MV型渦輪螺旋槳發動機 功率 4×11,000千瓦（14,800馬力） 性能數據 最大速度 925公里／時（500節，575哩／時） 爬升率 10米／秒（2,000呎／分） 最大升限 12,000米（39,000呎） 最大航程 15,000公里（8,100海浬，9,400哩） 翼負荷 606公斤／平方米（124磅／平方呎） 推重比 235 W／公斤（0.143馬力／磅） 武器装备 機炮 1具/2具AM-23型23公釐雷達控制機砲（機尾） 飛彈 空对地导弹： Kh-20、Kh-22 Kh-26、Kh-55 炸彈 60枚FAB-250炸彈，或 30枚FAB-500炸彈。 其他 载弹量（包括飞弹）：最多15,000公斤（33,000磅） Tu-95熊式MR型機 图-95 （俄语： Ту–95 ， 英语： Tu-95 ），北約代號： Bear （ 熊 ），是蘇聯圖波列夫設計局所研製，是全世界唯一服役的大型四渦輪螺旋槳發動機...
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