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

1. ## [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?

2. ## 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 $M_m = 6.42 \times 10^{23}$ ...

$x = 4.2 \times 10^7 \, meters$

3. ## 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

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

Originally Posted by topsquark
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.

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?

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

Originally Posted by SDF
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 ...

$\dfrac{a}{b} = \dfrac{c}{d-e} \implies ad - ae = bc$

7. ## 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:

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

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

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

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

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

8. ## 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 $(\sqrt{M_e} + \sqrt{M_m})$ ...

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