For a spherical celestial object of radius R , the acceleration due to gravity g

Question

For a spherical celestial object of radius R, the acceleration due to gravity g at a distance x from the center of the object is g = g0R2/x2, where g0 is the acceleration due to gravity at the object's surface and x > R. For the moon, take g0 = 1.63 m/s2 and R = 3200 km. If a rock is released from rest at a height of 5R above the lunar surface, with what speed does the rock impact the moon? Hint: Its acceleration is a function of position and increases as the object falls. So do not use constant acceleration free-fall equations, but go back to basics.

In km/s

Explanation / Answer

The acceleration is the second derivative of the position with respect to time, right?, So:

a = d^2xdt^2 = -(g0*R^2)/x2

(The negative sign is because the direction of acceleration is downward, toward the center of the moon)

This is a nonlinear second order differential equation. Notice that the independent variable, t, does not appear explicitly. In such cases, it turns out that making the substitution, v = dx/dt will usually simplify the equation.

Define v = dx/dt

then d^2xdt^2 = dv/dt = dv/dx * dx/dt = dv/dx * v

Substituting v*dv/dx for d^2xdt^2 in the original equation we get:

v*dv/dx = -(g0*R^2)/x2

This is now a separable first order equation, and even better, it gives the speed of the rock (v = dx/dt) as a function of position, which is exactly what we need to find.

Separating variables, we get:

v dv = -(g0*R^2) * dx/(x^2)

(1/2)*(v^2 - v0^2) = -(g0*R^2)*(1/x0 - 1/x)

v^2 = v0^2 - 2*(g0*R^2)*(1/x0 - 1/x)

We know that initially, v0 (the initial speed) is zero, and that the rock is dropped from a distance 6R from the center of the moon's center (5R above the surface of the moon), and we are interested in the value of v when x = R:

v^2 = -2*(g0*R^2)*(1/(6R) - 1/R)

v^2 = 2*(g0*R^2)*(5/6R)

v^2 = 10*g0*R/6

Plugging in the values for g0 and R, we get:

v^2 = 10*(1.63 m/s^2)*(3.2*10^6 m)/6

v^2 = 8.693*10^6 (m/s)^2

v = 2948 m/s

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