The following expression for the acceleration due to gravity works well for obje

Question

The following expression for the acceleration due to gravity works well for objects near the earths surface (G is the gravitational constant, Me is the mass of the earth, and are is the radius of the earth): g= GM/R^2. Use the above equation to calculate the gravitational acceleration at an altitude of 1000,000 meters above the earth. By what percentage is this acceleration different from that on the earths surface?

(2) how does air resistance alter the way we perceive falling objects?

(3) using the universal law of gravitation, show the gravitational acceleration experienced by an object independent of its mass.

Explanation / Answer

We have M (mass of earth) = 6 * 10^24 kg

and G = 6.67 * 10^-11

then g = G * M / R^2

take R = 1000000 m = 10^7 m

the g = ( 6.67 * 10^-11 * 6 * 10^24) / (10^7)^2

= 40.02 * 10^ 13 / 10^14 = 4 .002 m/sec^2 which as you can understand is much lesser than g on surface = 9.8 m/s^2

(b) Air resistance provides a backward drag to object. More the surface area of the object, more is the drag on this object. hence it cancels a little of the effect of gravitation. Hemce our perception is that the object must be falling slowly. This is why a sack of cotton and a sack of iron of same mass will not fall at the same rate and time, becausethe drag will be more on the sack of iron.

(c) We have F = G * m1 * m2 / R^2

If F is the force on a body of mass m1

the acceleration due to that force = force / mass = G * m1 * m2 / (R^2 * m1) = G * m2 / R^2

Hence the acceleration due to gravity is independent of mass.

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