Hello from MrBillDoesMath!
Answer:
The y axis
Discussion:
The equation is
F = Β (G (m1)(m2)) / r^2
so the question is equivalent to asking what is the asymptote of 1/r^2 as r approaches 0, The asymptote is the y axis as shown in the attachment where 1/r^2 is graphed.
Thank you,
MrB
Answer with explanation:
The force of gravity between two objects is given by
Β Β [tex]F_{g}=\frac{-Gm_{1}m_{2}}{r^2}[/tex]
Replacing , force by , y and product of -G Β the gravitational constant, masses [tex]m_{1}, m_{2}[/tex] by t,the above equation reduces to,
Β [tex]y=\frac{t}{x^2}[/tex]
Horizontal Asymptote, is, y=0
Β [tex]y= \lim_{x \to \infty} f(x)\\\\y = \lim_{x \to \infty} \frac{1}{x^2}\\\\y=0[/tex]
And Vertical Asymptote is, x=0
Β [tex]x= \lim_{y \to \infty} f(y)\\\\x = \lim_{y \to \infty} \frac{-1}{\sqrt{y}}\\\\x=0[/tex]
The meaning of vertical asymptote in context of the graph of the function is Β ,that when Distance between two objects reduces to Zero, the force of attraction between two bodies, reduces to zero.
