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When you drive a .36 kg apple, Earth exerts a force on It that

  1. Feb 18, 2012 #1
    When you drive a .36 kg apple, Earth exerts a force on It that accelerates it at 9.8 m/s squared toward the earths surface. According to newtons third law the Apple must exert an equal but opposite force on earth. If the mass of the earth is 5.89x10^24 kg,what is the magnitude of the earth's acceleration toward the apple?

    I have tried this problem a couple of times and get it wrong. My math is: .36*9.8/5.98x10^24.
     
  2. jcsd
  3. Feb 18, 2012 #2
    Re: Physics

    Hello KellySierra22,
    Instead of giving your workings with numbers, it is easier to understand the problem in algebra.

    We start with the well known F=ma, however we must be careful about which force and which mass (although the acceleration is constant). So I label them

    [itex] F_{apple \rightarrow earth}= m_{apple} \times g_{apple \rightarrow earth} [/itex]

    That is the force the earth exerts on the apple is the mass of the apple times acceleration, you must calculate this force first. Now that you have done so we can use Newtons third law

    [itex] F_{apple \rightarrow earth}=F_{earth \rightarrow apple}[/itex]

    It may help to draw this pictorially. So you now have an equation relating the force you already have to the force the apple exerts on the earth. You can now calculate the acceleration on the earth from the apple using

    [itex] F_{earth \rightarrow apple}= m_{earth} \times g_{earth \rightarrow apple} [/itex]

    You must solve for the acceleration of the earth towards the apple.
     
  4. Feb 18, 2012 #3

    gneill

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    Staff: Mentor

    Re: Physics

    Hello KellySierra22, Welcome to Physics Forums.

    The expression that you have shown should give you the numerical value for the acceleration of the Earth towards the apple. What value did you obtain when you completed the math?
     
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