Ian: New page: $F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$ where $F$ is the force on the particle, ${{m}_{1}}$ is the mass of the moon, ${{m}_{2}}$ is the ...

Ian — Mon, 30 Mar 2009 14:11:53 GMT

New page: <math>F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}</math> where <math>F</math> is the force on the particle, <math>{{m}_{1}}</math> is the mass of the moon, <math>{{m}_{2}}</math> is the ...

Ny sida

<math>F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}</math>

where
<math>F</math>
is the force on the particle,
<math>{{m}_{1}}</math>
is the mass of the moon,
<math>{{m}_{2}}</math>
is the mass of the particle and
<math>d</math>
is the radius of the moon.

As
<math>F={{m}_{2}}a</math>
where
<math>a</math>
is the acceleration of the particle

<math>a=\frac{G{{m}_{1}}}{{{d}^{2}}}</math>

As
<math>G=6.67\times {{10}^{-11}}\text{ k}{{\text{g}}^{\text{-1}}}{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{-2}}}</math>
we get

<math>\begin{align}
& a=\frac{6.67\times {{10}^{-11}}\text{ k}{{\text{g}}^{\text{-1}}}{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{-2}}}\times \text{7}.\text{38}\times \text{1}{{0}^{\text{22}}}\text{kg}}{{{\left( \text{1}.\text{73}\times \text{1}{{0}^{\text{6}}}\text{m} \right)}^{2}}} \\
& =\frac{6.67\times \text{7}.\text{38}}{\text{1}.\text{7}{{\text{3}}^{2}}\times 10}\text{m}{{\text{s}}^{-2}}=1.64\text{m}{{\text{s}}^{-2}} \\
\end{align}</math>

Solution 2.7 - Versionshistorik

Ian: New page: F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}} where F is the force on the particle, {{m}_{1}} is the mass of the moon, {{m}_{2}} is the ...

Ian: New page: $F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$ where $F$ is the force on the particle, ${{m}_{1}}$ is the mass of the moon, ${{m}_{2}}$ is the ...