7.4.2 Gravitational Potential of a Point Mass

Consider a massive astronomical body of mass M and a tiny satellite of mass m, separated by distance r.

We can say that M and m as a system has a GPE of \displaystyle U=-\frac{{GMm}}{r}.

Alternatively, we can say that M’s gravitational field is producing a gravitational potential f around it, thus causing m to have a GPE of  U=\phi m.

Since gravitational potential is GPE per unit mass, the gravitational potential produced by M is

\displaystyle \begin{aligned}\phi &=-\frac{{GMm}}{r}\div m\\&=-\frac{{GM}}{r}\end{aligned}

A gravitational potential landscape can be represented by a set of equipotential lines, where each line joins all the points with the same gravitational potential (much like the contour lines you learnt in geography). The gravitational potential produced by M can thus be represented by a potential map consisting of concentric circles centred about M.

We derived this formula assuming that M is a point mass (the assumption is implicit when we use the formula \displaystyle U=-\frac{{GMm}}{r}). Massive astronomical bodies (stars, planets, etc) are of course not point masses. But they are spherical in shape and have a spherical symmetry, and can be shown mathematically to be equivalent to a point mass situated at the centre of the sphere to someone outside the sphere. So this formula is valid for stars and planets as well, but for points outside the spherical surface only.

Our very own Earth can be modelled as a uniform sphere of radius  {{R}_{E}}=6370\text{ km} and mass {{M}_{E}}=5.97\times {{10}^{{24}}}\text{ kg}. The theoretical values of \displaystyle \phi =-\frac{{G{{M}_{E}}}}{r} for r\ge {{R}_{E}} can thus be calculated.

A few important points:

  • f is a scalar quantity (unlike g).
  • f is always negative.
  • f is inversely proportional to r, and approaches zero at the point at infinity.

Example

The graph below shows the total gravitational potential along the line between planet A and moon B.

a) Describe the significance of position Y.

b) Calculate the minimum speed a projectile must be launched from the surface of moon B in order for it to arrive at planet A. (Ignore effects of air resistance)

c) Calculate the minimum landing speed on planet A.

Solution

a)

Y is the gravitational null point where the resultant gravitational field strength is zero. On the left of Y, the gravitational field is towards planet A. On the right of Y, the gravitational field is towards moon B.

b)

The projectile must start with enough KE to arrive at Y. (Upon reaching Y, the satellite can just fall to Earth)

By PCOE:

\displaystyle \begin{aligned}(KE+GPE)\text{ at Z}&=(KE+GPE)\text{ at Y}\\\frac{1}{2}m{{v}^{2}}+m{{\phi }_{Z}}&=0+m{{\phi }_{Y}}\\\frac{1}{2}{{v}^{2}}+(-35.0\times {{10}^{6}})&=-22.5\times {{10}^{6}}\\v&=5000\text{ m }{{\text{s}}^{{-1}}}\end{aligned}

c)

By PCOE:

\displaystyle \begin{aligned}(KE+GPE)\text{ at Z}&=(KE+GPE)\text{ at X}\\\frac{1}{2}m{{v}_{Z}}^{2}+m{{\phi }_{Z}}&=\frac{1}{2}m{{v}_{X}}^{2}+m{{\phi }_{X}}\\\frac{1}{2}{{(5000)}^{2}}+(-35.0\times {{10}^{6}})&=\frac{1}{2}{{v}_{X}}^{2}-102.5\times {{10}^{6}}\\{{v}_{X}}&=12649\text{ m }{{\text{s}}^{{-1}}}\end{aligned}

Video Explanation

What is Gravitational Potential Energy?

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s