Option 3 : \(\frac{1}{2}\)
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- Consider the mass of earth as ‘M’ and radius of earth ‘R’.
- Now we consider a planet twice the mass and radius of earth as 2M and 2R.
- We know the earth g = GM/R2 i.e. is g\(\; \propto \frac{M}{{{R^2}}}\) where G is Universal Gravitational Constant.
- Let g1 is acceleration due to gravity on earth and g2 is acceleration due to the gravity of another planet.
- So g2/g1 = (2M/M) × (R/2R)2
- c
- According due to gravity on surface of planet is n times of earth we get g2 = g1(n)
- Hence, the value n is ½.
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Text Solution
Solution : Let the acceleration due to gravity of that planet be `g_(p)` <br> As, `g_(p)=(GM_(p))/(R_(p)^(2))` <br> According to the question <br> `M_(p)=2M_(e)` and `R_(p)=2R_(e)` <br> `therefore g_(p)=(G2M_(e))/(4R_(e)^(2))` <br> `therefore g_(p)=(1)/(2)g_(e) " " .....[because g_(p)=(GM_(e))/(R_(e)^(2))]`