You are viewing an old version of this page. View the current version.

Compare with Current View Page History

« Previous Version 14 Next »

The gravitational force exerted by the earth on an object near the earth's surface.

The Force of Gravity Near Earth's Surface

Defining "Near"

Suppose an object of mass m is at a height h above the surface of the earth. Assume that the earth is spherical with radius RE. Working in spherical coordinates with the origin at the center of the earth, the gravitational force on the object from the earth will be:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ \vec

Unknown macro: {F}

= - G \frac{M_

Unknown macro: {E}

m}{(R_

+h)^{2}} \hat

Unknown macro: {r}

]\end


A Taylor expansion gives:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ \vec

Unknown macro: {F}

\approx - G \frac{M_

Unknown macro: {E}

m}{R_

^{2}}\left(1 - 2\frac

Unknown macro: {h}

{R_{E}} + ...\right)\hat

Unknown macro: {r}

]
\end


Thus, for h/RE << 1, the gravitational force from the earth on the object will be essentially independent on altitude above the earth's surface and will have a magnitude equal to:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ F_

Unknown macro: {g}

= mG\frac{M_{E}}{R_

Unknown macro: {E}

^{2}} ]\end

Defining g

The above expression is of the form:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ F_

Unknown macro: {g}

= mg ]\end


if we take:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ g = G\frac{M_{E}}{R_

Unknown macro: {E}

{2}} = \left(6.67\times 10{-11}\mbox

Unknown macro: { N}

\frac{\mbox

Unknown macro: {m}

^{2}}{\mbox

Unknown macro: {kg}

{2}}\right)\left(\frac{5.98\times 10

Unknown macro: {24}

\mbox{ kg}}{(6.37\times 10^

Unknown macro: {6}

\mbox

Unknown macro: { m}

)^{2}}\right) = \mbox

Unknown macro: {9.8 m/s}

^

Unknown macro: {2}

]\end

Gravitational Potential Energy Near Earth's Surface

Near the earth's surface, if we assume coordinates with the +y direction pointing upward, the force of gravity can be written:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ \vec

Unknown macro: {F}

= -mg \hat

Unknown macro: {y}

]\end

Since the "natural" ground level varies depending upon the specific situation, it is customary to specify the coordinate system such that:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ U(0) \equiv 0]\end

The gravitational potential energy at any other height y can then be found by choosing a path for the work integral that is perfectly vertical, such that:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ U = U(0) - \int_

Unknown macro: {0}

^

Unknown macro: {y}

(-mg)\;dy = mgy]\end

For an object in vertical freefall (no horizontal motion) the associated potential energy curve would then be:

For movement under pure near-earth gravity, then, there is no equilibrium point. At least one other force, such as a normal force, tension, etc., must be present to produce equilibrium.

Example Problems involving Near-earth Gravity

ExampleProblemsinvolvingGravitationalForce"> Example Problems involving Gravitational Force

Error formatting macro: contentbylabel: com.atlassian.confluence.api.service.exceptions.BadRequestException: Could not parse cql : null

ExampleProblemsinvolvingGravitationalPotentialEnergy"> Example Problems involving Gravitational Potential Energy

Error formatting macro: contentbylabel: com.atlassian.confluence.api.service.exceptions.BadRequestException: Could not parse cql : null
  • No labels