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Solution 2.8

From Mechanics

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(New page: Consider a particle on the surface of the planet. <math>F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}</math> where <math>F</math> is the gravitational force on a particle on the surface of t...)
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Revision as of 15:02, 12 March 2010

Consider a particle on the surface of the planet.

F=d2Gm1m2

where F is the gravitational force on a particle on the surface of the planet, in other words, its weight.

m1 is the mass of the planet, m2 is the mass of the particle and d is the radius of the planet.

As F=m2a where a is the acceleration of the particle very close to the surface of the planet,


a=d2Gm1

Thus

d2=aGm1

As G=6.6710−11 kg-1m3s-2 we get


d2=3.26.6710−1151020=3.233.35109=10.4109=1.041010d=1.04105=1.02105 m=102 km

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