Results 1 to 8 of 8
Like Tree8Thanks
  • 2 Post By skeeter
  • 1 Post By topsquark
  • 1 Post By SDF
  • 1 Post By skeeter
  • 3 Post By skeeter

Thread: [Physics] Net force of gravity on an object

  1. #1
    SDF
    SDF is offline
    Member SDF's Avatar
    Joined
    Apr 2014
    From
    Canada
    Posts
    75
    Thanks
    5

    [Physics] Net force of gravity on an object

    I'm stuck on a physics problem involving the net force of gravity on an object x kilometers away from Earth. The correct answer is 42000 km though I cannot seem to get it when I do the calculations. If someone could tell me where I am going wrong I would appreciate it. Note: I've tried several other problems just like this (albeit with different masses) so I don't think it's just a minor error that is causing my calculations to be way off - rather an error in the logic.

    Here is the problem, along with my work:


    Take mass of Earth ME = 6.00 x 10^24 kg; mass of Mars MM = 6.42 x 10^23 kg; and the distance between their centres REM = 5.57 x 10^7 m. At what point (in km from Earth) between Earth and Mars is the net force of gravity on a body by both Earth and Mars exactly zero?

    Let x = the distance of the object from earth
    Let m = the mass of the object

    $${F_E} = \frac{G\cdot {M_E}\cdot m}{x^2} = \frac{(6.674 x 10^{-11})(6.00 x 10^{24})(m)}{x^2}$$

    $${F_M} = \frac{G\cdot {M_M}\cdot m}{(5.57 x 10^7- x)^2} = \frac{(6.674 x 10^{-11})(6.42 x 10^{23})(m)}{(5.57 x 10^7 - x)^2}$$

    Set ${F_E} = {F_M}$ ... I have no idea why the TeX below doesn't work as it's literally the same code copied from the above.

    $$
    \frac{(6.674 x 10^{-11})(6.00 x 10^{24})(m)}{x^2} = \frac{(6.674 x 10^{-11})(6.42 x 10^{23})(m)}{(5.57 x 10^7 - x)^2}$$


    Divide out the gravitational constant (6.674 x 10^-11), the object mass m, and the 10^24 and 10^23.

    $$\frac{6}{x^2} = \frac{6.42}{(5.57 x 10^7 - x)^2}$$

    $$(\frac{(5.57 x 10^7 - x)}{x})^2 = \frac{6.42}{6}$$

    Take square root of both sides and I'm left with:

    $$\frac{55700000 - x}{x} = 1.0344$$

    Multiply both sides by x to get rid of it in the denominator on the left:

    $$55700000 - x = 1.0344x$$

    Add x to both sides, then divide to get x on its own:

    $$x = 27379079.82$$

    The answer I get (2.73 x 10^6) is way off and I don't know why. I've tried multiple times to isolate x in different ways but I always end up with a massive number instead of the 42000 km that is the answer. Any idea why?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,154
    Thanks
    3672

    Re: [Physics] Net force of gravity on an object

    Take mass of Earth ME = 6.00 x 10^24 kg; mass of Mars MM = 6.42 x 10^23 kg; and the distance between their centres REM = 5.57 x 10^7 m. At what point (in km from Earth) between Earth and Mars is the net force of gravity on a body by both Earth and Mars exactly zero?
    $\dfrac{GM_e m}{x^2} = \dfrac{GM_m m}{(R-x)^2}$

    $\dfrac{M_e}{x^2} = \dfrac{M_m}{(R-x)^2}$

    $\dfrac{\sqrt{M_e}}{x} = \dfrac{\sqrt{M_m}}{R-x}$

    $R\sqrt{M_e} - x \sqrt{M_e} = x\sqrt{M_m}$

    $R\sqrt{M_e} = x \sqrt{M_e} + x\sqrt{M_m}$

    $R\sqrt{M_e} = x (\sqrt{M_e} + \sqrt{M_m})$

    $\dfrac{R\sqrt{M_e}}{\sqrt{M_e} + \sqrt{M_m}} = x$

    substitute in $R = 5.57 \times 10^7$, $M_e = 6.00 \times 10^{24}$, and $M_m = 6.42 \times 10^{23}$ ...

    $x = 4.2 \times 10^7 \, meters$
    Thanks from topsquark and SDF
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,071
    Thanks
    708
    Awards
    1

    Re: [Physics] Net force of gravity on an object

    SDF's signature:
    The spam is sanctioned. ;-)
    I earned that avatar buddy so yeah, keep thinking that way.

    -Dan
    Thanks from skeeter
    Follow Math Help Forum on Facebook and Google+

  4. #4
    SDF
    SDF is offline
    Member SDF's Avatar
    Joined
    Apr 2014
    From
    Canada
    Posts
    75
    Thanks
    5

    Re: [Physics] Net force of gravity on an object

    Quote Originally Posted by topsquark View Post
    SDF's signature:

    I earned that avatar buddy so yeah, keep thinking that way.

    -Dan
    Haha I'd forgotten about that. I've changed it just for you.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  5. #5
    SDF
    SDF is offline
    Member SDF's Avatar
    Joined
    Apr 2014
    From
    Canada
    Posts
    75
    Thanks
    5

    Re: [Physics] Net force of gravity on an object

    Sorry to revive the thread, I thought I was clear on everything. Could you please clarify how you reached your fourth step?

    I understand up until here:

    $$\frac{\sqrt{M_{e}}}{x} = \frac{\sqrt{M_{m}}}{R-x}$$

    I'm not sure how you got this though:

    $$R\sqrt{M_e} - x \sqrt{M_e} = x\sqrt{M_m}$$

    My guess is you multiplied both sides by R and x so that they would cancel on their respective sides of the equation, however when I do that I get this:

    $$R\sqrt{M_{e}} = \frac{x\sqrt{M_{m}}}{-x}$$

    Where does the extra $$\sqrt{M_{e}}$$ come from?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,154
    Thanks
    3672

    Re: [Physics] Net force of gravity on an object

    Quote Originally Posted by SDF View Post
    Sorry to revive the thread, I thought I was clear on everything. Could you please clarify how you reached your fourth step?

    I understand up until here:

    $$\frac{\sqrt{M_{e}}}{x} = \frac{\sqrt{M_{m}}}{R-x}$$

    I'm not sure how you got this though:

    $$R\sqrt{M_e} - x \sqrt{M_e} = x\sqrt{M_m}$$
    cross multiply ...

    $\dfrac{a}{b} = \dfrac{c}{d-e} \implies ad - ae = bc$
    Thanks from SDF
    Follow Math Help Forum on Facebook and Google+

  7. #7
    SDF
    SDF is offline
    Member SDF's Avatar
    Joined
    Apr 2014
    From
    Canada
    Posts
    75
    Thanks
    5

    Re: [Physics] Net force of gravity on an object

    Ah ok, but why? If the goal is to isolate x why is that step necessary? I don't mean to be pedantic but I really don't understand the reason why that step is on the right track while simply trying to isolate x like I did isn't.

    This is how I proceeded:

    \frac{\sqrt{M_{e}}}{x} = \frac{\sqrt{M_{m}}}{R-x}

    R\sqrt{M_{e}} = \frac{x\sqrt{M_{m}}}{-x}

    R\sqrt{M_{e}}=-x \sqrt{M_{m}}

    \frac{R\sqrt{M_{e}}}{\sqrt{M_{m}}} = -x

    x=\frac{-R\sqrt{M_{e}}}{\sqrt{M_{m}}}
    Last edited by SDF; Oct 28th 2017 at 10:46 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,154
    Thanks
    3672

    Re: [Physics] Net force of gravity on an object

    okay ... one more time with a full explanation for each step

    $\dfrac{\sqrt{M_e}}{x} = \dfrac{\sqrt{M_m}}{R-x}$

    multiply both sides by the common denominator $x(R-x)$ ...

    $x(R-x) \cdot \dfrac{\sqrt{M_e}}{x} = x(R-x) \cdot \dfrac{\sqrt{M_m}}{R-x}$

    since $x \ne 0$ and $x \ne R$, the common factors in the numerator and denominator divide out ...

    $\cancel{x}(R-x) \cdot \dfrac{\sqrt{M_e}}{\cancel{x}} = x(\cancel{R-x}) \cdot \dfrac{\sqrt{M_m}}{\cancel{R-x}}$

    with me so far? ... rewrite the equation sans the cancelled factors

    $(R-x)\sqrt{M_e} = x\sqrt{M_m}$

    distribute the left side of the equation ...

    $R\sqrt{M_e} - x\sqrt{M_e} = x\sqrt{M_m}$

    add $x\sqrt{M_e}$ to both sides ...

    $R\sqrt{M_e} = x\sqrt{M_e} + x\sqrt{M_m}$

    factor out the common factor $x$ from both terms on the right side of the equation ...

    $R\sqrt{M_e} = x(\sqrt{M_e} + \sqrt{M_m})$

    divide both sides by $(\sqrt{M_e} + \sqrt{M_m})$ ...

    $\dfrac{R\sqrt{M_e}}{\sqrt{M_e} + \sqrt{M_m}} = x$
    Thanks from topsquark, romsek and SDF
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. force on a submerged object?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Aug 4th 2013, 08:50 AM
  2. Replies: 2
    Last Post: Nov 18th 2010, 09:22 AM
  3. physics -gravity
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: Oct 30th 2010, 08:00 AM
  4. Replies: 2
    Last Post: Oct 6th 2010, 05:21 AM
  5. gravity force
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: Dec 22nd 2006, 10:45 PM

/mathhelpforum @mathhelpforum