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Question

The minimum and maximum distances of a satellite from the center of the Earth are 2R and 4R respectively, where R is the radius of Earth and M is the mass of Earth. The radius of curvature at the point of minimum distance is


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Solution

Step 1: Formula used

By conservation of angular momentum,

mv1R1=mv2R2v1R1=v2R2v1(2R)=v2(4R)v1=2v2v2=v12(1)

R1 Minimum distance of satellite from the center of earth

R2 Maximum distance of satellite from the center of earth

m Mass of the satellite

v1 Velocity of satellite at minimum distance

v2 Velocity of satellite at maximum distance

Step 2: Use conservation of energy and substitute the values

By conservation of energy

Total energy = Kinetic energy + Potential energy

(Kineticenergy+Potentialenergy)atmindistance=(Kineticenergy+Potentialenergy)atmaxdistance

-GMm2R+12mv12=-GMm4R+12mv22(2)

Substitute (1) in (2), we get

v1=GM6Rv2=2GM3R

Step 3: Find the radius of curvature

If r is the radius of curvature at the minimum distance,

mv12r=GMm(2R)2r=4v12R2GMr=8R3(Substitutingvalueofv1)

The radius of curvature at the point of minimum distance is r=8R3.


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