I just had this idea of orbiting a planet just by jumping and then flying upon it on its orbit kind of like superman. So,

Would it be theoretically possible or is there a chance of that small body to be & remain its unity?



Let's assume mass of the person plus spacesuit to be $m_1$=100kg

Asteroid density: $\rho=$2g/cm$^3$ (source) that is 2 000kg/m$^3$

15km/hour is a good common run. That's roughly v=4m/s

The orbital height is negligible comparing to the radius, assume 0 over surface.

Linear to angular velocity (1):
$$
\omega = {v \over r }
$$
Centripetal force (2):
$$
F = m r \omega ^2  
$$
Gravity force (3):
$$
F= G \frac{m_1 m_2}{r^2}
$$
Volume of a sphere (4):
$$
V = \frac{4}{3}\pi r^3 
$$
Mass of a sphere (5):
$$
m_2 = V \rho = \frac{4}{3}\pi r^3 \rho
$$
Combining (1),(2),(3), reducing:
$$
{ m_1 r v^2 \over r^2 } = G { m_1 * m_2 \over r^2 }
$$
$$
 r v^2 = G m_2 
$$
Combining with (5)
$$
 r v^2 = G \frac{4}{3}\pi r^3 \rho
$$

$$
 r^2 = \frac{v^2}{\rho G \frac{4}{3}\pi} 
$$

$$
 r = v ({\frac{4}{3}\pi G \rho})^{-{1 \over 2}}
$$
Substituting values:
$$
 r = 4 ({1.33333*3.14159* 6.67300*10^{-11} * 2000})^{-{1 \over 2}}  
$$

That computes to roughly 5.3 kilometers

More interestingly, the radius is directly proportional to the velocity,

$$
r[m] = 1337[s] * v [m/s] = 371.51[h/1000] * v[km/h] = 597[m*h/mile] * v[mph]
$$

So, a good walk on a 2km radius asteroid will get you orbiting.

Something to fit your bill would be Cruithne, a viable target for a space mission thanks to a very friendly orbit.

Note, while in rest on Cruithne, the astronaut matching the m_1=100kg would be pulled down with force of 4.5N while not in motion. That is like weighing about 450g or 1lbs on Earth.
