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Solution 2.10

From Mechanics

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Current revision (16:16, 3 February 2011) (edit) (undo)
 
Line 1: Line 1:
The Law of gravitation applied to a particle on Mars gives,
The Law of gravitation applied to a particle on Mars gives,
-
<math>F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}</math>
+
<math>F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{\, 2}}}</math>
where
where
Line 19: Line 19:
is the acceleration of the particle on Mars,
is the acceleration of the particle on Mars,
-
<math>a=\frac{G{{m}_{1}}}{{{d}^{2}}}</math>
+
<math>a=\frac{G{{m}_{1}}}{{{d}^{\, 2}}}</math>
The radius of Mars must be expressed in SI units.
The radius of Mars must be expressed in SI units.

Current revision

The Law of gravitation applied to a particle on Mars gives,

F=d2Gm1m2

where F is the force on a particle on the surface of Mars, m1 is the mass of Mars, m2 is the mass of the particle and d is the radius of Mars.

As F=m2a where a is the acceleration of the particle on Mars,

a=d2Gm1

The radius of Mars must be expressed in SI units.

d=3.4106 m

and as G=6.6710−11 kg-1m3s-2 we get

a=3.4106m26.6710−11 kg-1m3s-2 6.421023kg=3.426.676.42 ms−2=3.7 ms−2

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