Question

A student is studying the potential energy change of a 50 kg object raised 110 km above Earth's surface. What will be the percentage error if he simply used the approximate relation ΔU = mgΔy? Hint

Answer 1

Answer:

The percentage error is given by 99.9 %

Explanation:

Given:

Mass of object $m = 50$ kg

Height $h = 110$ km

From the formula of potential energy,

   $U = mgh$

Where $g = 9.8 \frac{m}{s}$

   $U = 50\times 9.8 \times 110000$

Here true value of potential energy,

   $U = 50 \times 9.8 \times 11 \times 10^{4}$

Approximate value of student,

   $U = 50 \times 9.8 \times 110$

Absolute error is given by

    = true value - approximate value

    = $50 \times 9.8 \times 11 \times 10^{4} - 50 \times 9.8 \times 110$

    = $53846100$

Hence percentage error,

    $= \frac{53846100}{50 \times 9.8 \times 11 \times 10^{4} }$

    $= 0.999 \times 100$ %

    $= 99.9$ %

Therefore, the percentage error is given by 99.9 %

Answer 2

This question involves the concepts of potential energy and variation in the value of acceleration due to gravity with altitude.

The percentage error if he uses the approximate relation will be "3.57 %".

First, we will calculate the change in potential energy from approximate relation:

$\Delta U_{approx.} = mg\Delta y$

where,

$\Delta U_{Approx.}$ = approximate change in potential energy = ?

m = mass of object = 50 kg

g = acceleration due to gravity on the surface of earth = 9.81 m/s²

Δy = height achieved = 110 km = 110000 m

Therefore,

$\Delta U_{approx.} = (50\ kg)(9.81\ m/s^2)(110000\ m)\\\Delta U_{approx.} = 53955000\ J$

Now, this is an approximate value because we used the value for acceleration due to gravity on the surface of the earth. But in actual it changes with the change in altitude. Hence, for actual change in potential energy we will calculate the value of acceleration due to gravity at the given altitude from the following formula:

$g_h = g(1-\frac{2h}{R})$

where,

g_h = acceleration due to gravity at given altitude = ?

R = radisu of the earth = 6.38 x 10⁶ m

Therefore,

$g_h=(9.81\ m/s^2)(1-\frac{2(110000\ m)}{6.38\ x\ 10^6\ m})\\g_h = 9.47\ m/s^2$

Hence, the actual change in potential energy will be:

$\Delta U = mg_h\Delta y = (50\ kg)(9.47\ m/s^2)(110000\ m)\\\Delta U = 52094482.8\ J$

Now, the percentage error will be:

$Error = \frac{\Delta U_{Approx.}-\Delta U}{\Delta U}*100\%\\\\Error = \frac{53955000\ J-52094482.8\ J}{52094482.8\ J}*100\%$

Error = 3.57 %

Learn more about variation in acceleration due to gravity with altitude here:

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