## Method

## The model

*R*and has mass

*M*.

The Earth is initially (

*t = 0*) at a distance of

*r*from the centre of the Sun and has mass

*m*. The radius of the Earth is negligible compared to the Sun.

*t*when it reaches the surface of the Sun at

*x = R*.

## The maths/physics bit

*G*is the universal constant of gravitation and \( x \) is the separation of the Sun and Earth. We use \( x \) rather than \( r \) because the distance will be varying in this problem and \( r \) is just the initial distance.

P.E. &=& \int^{r}_{\infty} \frac{GMm}{x^2} \textrm{d}x \\

& = & \left[ \frac{GMm}{x} \right]^r_{\infty} \\

& = & -\frac{GMm}{r}

\end{align} \]

*t*later during the fall the potential energy is given by: \[ P.E. = -\frac{GMm}{x}\] and the conservation of energy equation is \[\frac{1}{2}mv^2 -\frac{GMm}{x} = -\frac{GMm}{r} \]

*v*the subject: \[ \begin{align}

v^2 & = & 2GM\left( \frac{1}{x} - \frac{1}{r} \right) \\

v^2 & = & 2GM\left( \frac{r-x}{rx} \right) \\

\therefore v & = & -\left[ 2GM\left( \frac{r-x}{rx}\right) \right]^{\frac{1}{2}}

\end{align} \]

r &=& 1.5\times 10^{11} \textrm{m} \\

M &=& 2\times 10^{30} \textrm{kg} \\

R &=& 6.96\times 10^{8} \textrm{m} \\

G &=& 6.67\times 10^{-11} \frac{m^2}{kg s^2}

\end{align} \]

*v*can be expressed as a derivative: \[ \frac{\textrm{d}x}{dt} = -\left[ 2GM\left( \frac{r-x}{rx}\right) \right]^{\frac{1}{2}} \]

*T*has reached the surface of the Sun at

*R*.

*k*earlier the angles are 0 and \(\sin^{-1} \sqrt{\frac{r-R}{r}}\).

## The answer (and a detour to the Oort Cloud)

**Starting from rest, the Earth will take just under 64 days to fall into the Sun. It will plunge into the Sun at speed of 384 miles per second (roughly 1.4 million miles per hour).**