Radius of orbit = 26570801.43 m Determine its speed = 3875.33 m/s Fractional cha

Question

Radius of orbit = 26570801.43 m

Determine its speed = 3875.33 m/s

Fractional change in frequency due to time dilation = -0.000000000083434

(d) The gravitational "blueshift" of the frequency according to general relativity is a separate effect. It is called a blueshift to indicate a change to a higher frequency. The magnitude of that fractional change is given by deltaF/F = delta(Ug) / mc^2

where Ug is the change in gravitational potential energy of an object%u2013Earth system when the object of mass m is moved between the two points where the signal is observed. Calculate this fractional change in frequency due to the change in position of the satellite from the Earth's surface to its orbital position.

(e) What is the overall fractional change in frequency due to both time dilation and gravitational blueshift?

Explanation / Answer

df/f = dU/mc^2

but since its given a circular orbit, the potential energy of the satellite with respect to earth doesnot change at any position since its evenly placed from earth at any position , and it goes around with same velocity too, which is its orbital velocity sqrt(GM/R)

fractional change in freq. due to its positional change = 0

and.

overall fractional change in frequency, due to time dilation and gravitational blueshift = = 0.834 * 10^-10 + 0

= 0.834 * 10^-10

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