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 %