To solve this problem we will apply the concepts related to gravity according to the Newtonian definitions. From finding this value we will use the linear motion kinematic equations to find the time. Our values are
Comet mass [tex]M = 1.0*10^{13} kg[/tex]
Radius [tex]r = 1.6km = 1600 m[/tex]
Rock was dropped from a height 'h' from surface = 1m
The relation for acceleration due to gravity of a body of mass 'm' with radius 'r' is
[tex]g = \frac{GM}{R^2}[/tex]
Where G means gravitational universal constant and M the mass of the planet
[tex]g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}[/tex]
[tex]g = 2.607*10^{-4} m/s^2[/tex]
Now calculate the value of the time
[tex]h = \frac{1}{2} gt^2[/tex]
[tex]t = \sqrt{\frac{2h}{g}}[/tex]
[tex]t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}[/tex]
[tex]t = 87.58s[/tex]
The time taken for the rock to reach the surface is t = 87.58s
The time taken for the rock to reach the surface is 87.58 s.
Given data:
The mass of sphere is, [tex]m = 1.0\times 10^{13} \;\rm kg[/tex].
The radius of sphere is, [tex]r = 1.6 \;\rm km =1600 \;\rm m[/tex].
The dropping height is, h = 1.0 m.
Apply the concepts related to gravity according to the Newtonian definitions. From finding this value we will use the linear motion kinematic equations to find the time.
The relation for acceleration due to gravity of a body of mass 'm' with radius 'r' is,
[tex]g = \dfrac{GM}{r^{2}} \\\\g = \dfrac{6.67 \times 10^{-11} \times 1 \times 10^{-13}}{(1600)^{2}} \\\\g=2.607 \times 10^{-4} \;\rm m/s^{2}[/tex]
Now, use the second kinematic equation of motion to find the time,
[tex]h=ut+\dfrac{1}{2}gt^{2} \\\\1.0=0 \times t+\dfrac{1}{2} \times 2.607 \times 10^{-4} \times t^{2} \\\\t = 87.58 \;\rm s[/tex]
Thus, the time required rock to hit the surface is 87.58 s.
Learn more about the kinematic equations of motion here:
https://brainly.com/question/14355103
