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प्रश्न
The kinetic energy of a revolving satellite (mass m) at a height equal to thrice the radius of the earth (R) is ______.
विकल्प
`"mgR"/8`
`"mgR"/16`
`"mgR"/2`
`"mgR"/4`
उत्तर
The kinetic energy of a revolving satellite (mass m) at a height equal to thrice the radius of the earth (R) is `underline("mgR"/8)`.
Explanation:
As we know that the kinetic energy of a revolving satellite at a height h above the earth's surface is given as
KE = `1/2 "mv"^2 = "GMm"/"2r" ...(because "v" = sqrt"GM"/"r" "for circular orbit")`
`= "GMm"/(2("R + h"))`
where, R is the radius of the earth.
Given, height of a satellite from the earth surface h = 3R and mass of the satellite = m
So, KE = `"GMm"/(2("R" + 3"R")) = "GMm"/(2("4R"))`
KE = `"GMm"/(8"R") xx "R"/"R"`
∴ GMm`"R"/(8"R"^2) = "gmR"/8 ...(therefore "g" = "GM"/"R"^2)`
So, the kinetic energy, KE = `"gmR"/8`
