The escape velocity on the surface of a planet is given by v = v(2GM/R), where M

Question

The escape velocity on the surface of a planet is given by
v = v(2GM/R), where M and R are the mass and radius of the planet, respectively, and G is Newton's gravitational constant. (The escape velocity is the velocity needed to refute the "What goes up must come down" maxim, neglecting air resistance). a)Write v in terms of the average mass density, ?, instead of M. b) Assuming that the average density of the earth is four times that of Juptier, and that the radius of Jupiter is 11 times that of the earth, what is v(Jupiter)/v(Earth)? The escape velocity on the surface of a planet is given by
v = v(2GM/R), where M and R are the mass and radius of the planet, respectively, and G is Newton's gravitational constant. (The escape velocity is the velocity needed to refute the "What goes up must come down" maxim, neglecting air resistance). a)Write v in terms of the average mass density, ?, instead of M. b) Assuming that the average density of the earth is four times that of Juptier, and that the radius of Jupiter is 11 times that of the earth, what is v(Jupiter)/v(Earth)?

Explanation / Answer

a)
escape velocity v_e =sq.root (2GM/R)   ..... (1)
average mass density =mass of planet/volume of planet
d =M/(4/3)pR^3
M =[(4/3)pR^3]d ..... (2) substitute the equation(2) in (1) ,we get escape velocity v_e =sq.root ((8/3)pR^2Gd ) ...... (3) this equation indicates ,v_e in terms of the average mass density ................................................................................................. b) we can write the below relation by using equation(3) (v_Jupiter/v_Earth) =sq.root[(R_J/R_E)^2(d_J/d_E)] where d_E =4d_J and R_J =11R_E these values substitute in above equation we get (v_Jupiter/v_Earth) =sq.root[(11R_E/R_E)^2(d_J/4d_J)]                               =5.5

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