Question

62.One projectile after deviating from its path startamoving round the earth in a cirular path of radiequal to nine times the radius of earth R. Its timeperiod will be :-(1)2π√R/√g (2) 27×2π√R/√g​

Answer 1

The time period of the projectile is  T = 27 × 2 π √ R/g

Explanation:

As given:

r = 9 R

Time period

T = 2 π √ r ^3/ G M

But   G M = g R^ 2

Time period

T = 2 π √ r ^3/ g R^ 2

T = 2 π √ ( 9 R )^ 3 /g R ^2

T = 2 π ( 9 ) ^3 / 2 √ R/ 9

T = 27 × 2 π √ R/g

Thus the time period of the projectile is  T = 27 × 2 π √ R/g

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