# Slope of Tangent

• October 14th 2010, 11:50 AM
RogueDemon
Slope of Tangent
The gravitational potential energy of the MIR space station, in joules, at a distance of x kilometers from the centre of the Earth, is described by the equation
$E = -\frac{(5.38)(10)^{19}}{x}$. Determine the slope of the tangent to the graph at the given value of x.

$x = (6.5)(10)^6 m$

$x = (6.5)(10)^6 m = 6500 km$

$m = \frac{E(6500 + 0.001) - E(6500)}{0.001}$

$m = \frac{(-\frac{(5.38)(10)^{19}}{(6500 + 0.001)}) -(-\frac{(5.38)(10)^{19}}{(6500)})}{0.001}$

$m = 1.273(10)^{12}$

Is this correct?
• October 14th 2010, 01:08 PM
CSM
$E = -(5.38)(10)^{19}* \frac{1}{x}$
$E' = (5.38)(10)^{19}* \frac{1}{x^2}$
With your x, this gives $1.27337278 \times 10^{12}$