Question

A spaceship of mass 8600 kg is returning to Earth with its engine turned off. Consider only the gravitational field of Earth. Let R be the distance from the center of Earth. In moving from position R1 = 7300 km to position R1 = 6700 km the kinetic energy of the spaceship increases by _____.

Answer 1

Answer:

$\Delta KE = 4.20\times 10^{13}\ J$

Explanation:

given,

mass of spaceship(m) = 8600 Kg

Mass of earth = 5.972 x 10²⁴ Kg

position of movement of space ship

R₁ = 7300 Km

R₂ = 6700 Km

the kinetic energy of the spaceship increases by = ?

Increase in Kinetic energy = decrease in potential energy

    $\Delta KE = GMm (\dfrac{1}{R_2}-\dfrac{1}{R_1})$

    $\Delta KE = GMm (\dfrac{R_1-R_2}{R_2R_1})$

    $\Delta KE = 6.67 \times 10^{-11}\times 5.972 \times 10^{24}\times 8600 (\dfrac{7300 - 6700}{7300 \times 6700})$

    $\Delta KE = 6.67 \times 10^{-11}\times 5.972 \times 10^{24}\times 8600 (\dfrac{600}{48910000})$

    $\Delta KE = 4.20\times 10^{13}\ J$