Simulating solar system with Newton's law I made a simulation in C++ with Newtons law and test it comparing the planets positions with the position from Solar system Calculator Don Cross (which I converted from JavaScript to  C++) 
http://cosinekitty.com/solar_system.html
What I do is every time step(usually 1 second but step 0.2 second is very similar to 10 seconds step) :

*

*Calculate accelaration ( $= $ newton forzes $\times$ deltatime)

*Update speed and positions

*Compare postions with results from Don Cross solar calculator alghorithms

But after 10 days of simulation I get this distance deviation (to the calculator from Don Cross) results :
$\mathrm{Mercury} \  4498.7 \ \mathrm{km}$
$\mathrm{Venus \  X} \ 1939.8  \ \mathrm{km}$
$\mathrm{Earth \  X} \ 10614.6 \ \mathrm{km}$
$\mathrm{Moon \ X} \ 7800.2 \ \mathrm{km}$
$\mathrm{Mars \  X} \ 445.2  \ \mathrm{km}$
$\mathrm{Ceres \ X} \ 129.5  \ \mathrm{km}$
$\mathrm{Pallas \ X} \ 432.4 \ \mathrm{km}$
$\mathrm{Juno \ X} \  151.4 \ \mathrm{km}$
$\mathrm{Vesta \ X} \  157.6 \ \mathrm{km}$
$\mathrm{Ida \ X} \ 73.6  \ \mathrm{km}$
$\mathrm{Gaspra} \  455.3 \ \mathrm{km}$
$\mathrm{9P/T1} \  241.5 \ \mathrm{km}$
$\mathrm{19P/B} \ 402.7 \ \mathrm{km}$
$\mathrm{67P/C-G} \  533.2 \ \mathrm{km}$
$\mathrm{81P/W2} \  110.7  \ \mathrm{km}$
$\mathrm{Jupiter} \ 172.3 \ \mathrm{km}$
$\mathrm{Saturn} \  261.2 \ \mathrm{km}$
$\mathrm{Uranus} \ 71.4  \ \mathrm{km}$
$\mathrm{Neptune} \   31.3 \ \mathrm{km}$
$\mathrm{Pluto \  X \ } \ 45.7 \ \mathrm{km}$
As you see some planets have little desviations and some bigger, so my question is: Can Newton's be accurate? or Don Cross solar system calculator is not? Or there is black matter in that region? Or what else? 
void CGravitator::CalcAceleration(double timeseconds){
unsigned int i,j,iend;
if (sunStatic)iend=m_np-1;
else iend=m_np;
for (i = 0; i < iend; i++) {
        m_planetas[i].aceleration.set(0,0,0);
        CVector3 totalGravitationalForce;                                       
        // Loop through all bodies in the universe to calculate their gravitational pull on the object (TODO: Ignore very far away objects for better performance)


     for (j = 0; j < m_np; j++) {
            if (i == j) continue; // No need to calculate the gravitational pull of a body on itself as it will always be 0.                                
            double distancia =CVector3::Distancia(m_planetas[i].pos,m_planetas[j].pos);
            double force = KGNEWTON * m_planetas[i].masa * m_planetas[j].masa / pow(distancia, 2);
            CVector3 forceDirection = CVector3::Normalize(m_planetas[j].pos - m_planetas[i].pos);
            
            totalGravitationalForce += forceDirection * force;
        }
            CVector3 incspeed = totalGravitationalForce / m_planetas[i].masa ;          
        m_planetas[i].aceleration += incspeed * timeseconds;
        
    
}

 A: You need to use a better numerical method. Euler’s method is notoriously bad for orbital mechanics because the numerical errors always accumulate. In particular, Euler’s method does not conserve energy, so you get orbits that just magically gain energy and spiral away out of control.
You need to use something like the Verlet method or some other symplectic integrator.
https://en.wikipedia.org/wiki/Verlet_integration
https://en.wikipedia.org/wiki/Symplectic_integrator
A: Besides the deficiencies in the numerical method pointed out in the other answer, simulating Mercury's orbit must take into account Jupiter's gravitation, as well as relativistic effects. This only explains a tiny fraction of the deviation but should be considered when the numerical method gets better.
Apparently the  perihelion precession of Mercury due to Jupiter's pull and relativistic effects is about 574 arcseconds in a century, or 1.57E-2 arcseconds/day. With an orbital circumference of about 3.6E8 km and an arcsecond being 1.296E-6 of a turn, that amounts to about 4.3km, or 43 km in 10 days.
While the difference in the position of the perihelion does not directly translate into a location difference it should give an idea of the effect.
