The minimum velocity needed for a body to escape from the earth is given by v = (square-root)(2GM/R) where M is the mass of the earth and R is its radius. Show that the equation is dimensionally correct. The dimensions of G are M^-1L^3T^-2
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F=Gm1m2/r^2
=>G=Fr^2/m1m2
G=N m^2/ kg^2
=Kg ms^-2 ×m^2 /kg^2
=m^3 kg^-1 s^-2
m is meter,s is second,kg is mass,(units in SI )
So
[G]=[L^3 M^-1 T^-2]
R is length,dimension is L,
M is mass, dimension is M
v=√2GM/R
[v]=[√GM/R]
=>[LT^-1]=[√(L^3 M^-1 T^-2 M)/L]
=>[LT^-1]=[√(L^2 T^-2 )]
=>[LT^-1]=[L T^-1]
=>G=Fr^2/m1m2
G=N m^2/ kg^2
=Kg ms^-2 ×m^2 /kg^2
=m^3 kg^-1 s^-2
m is meter,s is second,kg is mass,(units in SI )
So
[G]=[L^3 M^-1 T^-2]
R is length,dimension is L,
M is mass, dimension is M
v=√2GM/R
[v]=[√GM/R]
=>[LT^-1]=[√(L^3 M^-1 T^-2 M)/L]
=>[LT^-1]=[√(L^2 T^-2 )]
=>[LT^-1]=[L T^-1]