Respuesta :
The answer is easy if you know the physics.
a/a' = (r/R)^2 = (r/9r)^2 = 1/81; so that a = a'/81.
A's radial acceleration must be 1/81 of B's. And each acceleration is a = v^2/R and a' = V^2/r, where v and V are the tangential speeds you want the ratio for.
a = v^2/R = v^2/9r = V^2/81r = a'/81 In which case v^2/V^2 = 9/81 = 1/9; so that 1/3 = Va/Vb ANS
a/a' = (r/R)^2 = (r/9r)^2 = 1/81; so that a = a'/81.
A's radial acceleration must be 1/81 of B's. And each acceleration is a = v^2/R and a' = V^2/r, where v and V are the tangential speeds you want the ratio for.
a = v^2/R = v^2/9r = V^2/81r = a'/81 In which case v^2/V^2 = 9/81 = 1/9; so that 1/3 = Va/Vb ANS
Answer:
The ratio of speed will be 0.408
Explanation:
We have given that Planet A is 6 times farther than planet B
So [tex]R_A=6R_B[/tex]
We know that speed is given by [tex]v_A=\sqrt{\frac{GM}{R_A}}[/tex], here G is gravitational constant and [tex]R_A[/tex] is distance from star to planet A
As [tex]R_A=6R_B[/tex]
So [tex]v_A=\sqrt{\frac{GM}{6R_B}}[/tex]-----EQN 1
Speed of planet B [tex]v_B=\sqrt{\frac{GM}{R_B}}[/tex]------RQN 2
Dividing equation 1 by equation 2
[tex]\frac{v_A}{V_B}=\sqrt{\frac{1}{6}}[/tex]
[tex]\frac{v_A}{V_B}=0.408[/tex]
So the ratio of speed will be 0.408