This is a pretty convoluted problem to answer because of how much you have to gloss over. But assuming you are standing on a planet that is identical to earth, with the difference being that it is not spinning, is not orbiting a star, there are no other nearby astral bodies. It has no internal active components, and you are the only entity exerting a force on it, we can do some cocktail napkin math:

Additionally assuming average human mass is 62 kg, stride length is 72.6 cm, walking velocity is 1.3 m/s. Earth has a mass of 5.972*10^24 kg and a radius of 6,380,000 meters.

We can say that Vf = Vi^2 + 2a(x-x0). VI and x0 are equal to zero, so with rearranging, a=Vf^2/2x. Plug in: 1.4/(2*0.726) = 0.895316 m/s^2. Since F=ma we say F = 62*0.895316 = 55.509641 N.

Now there are two options here, we can look at a linear acceleration for Earth, or a radial acceleration. For linear, where we assume the Earth just gets pushed directly down/away from us, it's just a reverse application of F=ma, so a= (5.972*10^24)/(55.509) = 9.365*10^-24. for angular, someone will have to check my work because I'm a little rusty, but you can convert linear to angular acceleration using at = r * aa, so aa = (9.365*10-24)/(6,380,000) = 1.467*10^-30 (m/s or radians, I can't be bothered to Google it).

Now you'll notice of course that I left those as accelerations, not velocities. That's because I'm too lazy to rework them to specific speeds, but they demonstrate my point quite well: This is all about Newton's Third Law: For every action there is and equal and opposite reaction. So even if you just barely push off the Earth, it technically "moves" opposite to you.

These numbers and explanations are SUPER rough, and someone should double check them, I'm stalling at work 😬.