In the absence of air resistance, a man on the surface of the Earth can push a nucleus to $l = 22 m$. Can he push the same nucleus on the surface of an asteroid so that it becomes a satellite? The mass of the asteroid is $M = 3 \cdot 10^{13} t$, radius $R = 10 km$.
Decision:
The result of human impact on the nucleus is to impart to it some initial velocity $v$. A body thrown from the Earth's surface with initial velocity $v$ at an angle $\alpha$ to the horizon by the moment of falling travels in the horizontal direction a path $l$ defined by the formula
$l=\frac{v^{2}}{g} \sin(2 \alpha)$ ,(1)
where $g$ is the acceleration of free fall on the Earth. As follows from (1), the greatest range $l_{max}$ for a given velocity v is achieved at a throwing angle $\alpha = 45°$. In this connection, it is obvious that $l_{max}=v^{2}/g$ hence $v = \sqrt{gl_{max}}$. The greatest range of the kernel $s_{max}$ corresponds to the greatest value of the initial velocity that a person can give the kernel $v = v_{max}$. Doing the calculations, we find that $v_{max} = 15 m/s$. Let us denote by $v_{1}$ the first space velocity for the asteroid, i.e. $v_{1}$ is the minimum initial velocity that must be imparted to a body on the surface of an asteroid in order for it to become its satellite. If $v_{max} < v_{1}$, then a person cannot push the nucleus so that it becomes a satellite, if $v_{max} > v_{1}$, then they can. Let's find the value of $v_{1}$. Let us denote the mass of the nucleus by $m$. Obviously, it will revolve around the asteroid on a circle of radius $R$ if the condition is satisfied
$\frac{GmM}{R^{2}} = \frac{mv^{2}}{R}$.
Hence we get
$v_{1}=\sqrt{ \frac{GM}{R}}= 14 m/s$.
Thus, $v_{max} > v_{1}$, hence a person can push the nucleus so that it becomes a satellite of an asteroid.
Decision:
The result of human impact on the nucleus is to impart to it some initial velocity $v$. A body thrown from the Earth's surface with initial velocity $v$ at an angle $\alpha$ to the horizon by the moment of falling travels in the horizontal direction a path $l$ defined by the formula
$l=\frac{v^{2}}{g} \sin(2 \alpha)$ ,(1)
where $g$ is the acceleration of free fall on the Earth. As follows from (1), the greatest range $l_{max}$ for a given velocity v is achieved at a throwing angle $\alpha = 45°$. In this connection, it is obvious that $l_{max}=v^{2}/g$ hence $v = \sqrt{gl_{max}}$. The greatest range of the kernel $s_{max}$ corresponds to the greatest value of the initial velocity that a person can give the kernel $v = v_{max}$. Doing the calculations, we find that $v_{max} = 15 m/s$. Let us denote by $v_{1}$ the first space velocity for the asteroid, i.e. $v_{1}$ is the minimum initial velocity that must be imparted to a body on the surface of an asteroid in order for it to become its satellite. If $v_{max} < v_{1}$, then a person cannot push the nucleus so that it becomes a satellite, if $v_{max} > v_{1}$, then they can. Let's find the value of $v_{1}$. Let us denote the mass of the nucleus by $m$. Obviously, it will revolve around the asteroid on a circle of radius $R$ if the condition is satisfied
$\frac{GmM}{R^{2}} = \frac{mv^{2}}{R}$.
Hence we get
$v_{1}=\sqrt{ \frac{GM}{R}}= 14 m/s$.
Thus, $v_{max} > v_{1}$, hence a person can push the nucleus so that it becomes a satellite of an asteroid.