Question

Zero, a hypothetical planet, has a mass of 4.7 x 1023 kg, a radius of 3.0 x 106 m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial kinetic energy of 5.0 x 107 J, what will be its kinetic energy when it is 4.0 x 106 m from the center of Zero? (b) If the probe is to achieve a maximum distance of 8.0 x 106 m from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?

Answer 1

a)

Mass of planet = M = 4.7*10²³ kg

Radius of planet = r = 3 *10⁶ m

Mass of space probe = 10kg

Acceleration due to gravity at a height h on planet

$g=\frac{GM}{(r+h)^{2}}$

Total Energy of Space probe when launched from surface = Kinetic energy + Potential Energy

= 5.0*10⁷ J + 0                  (Assuming potential energy to be zero at surface)

Potential energy at an height r+h= 4 *10⁶ m from the center of planet,i.e. h=1*10⁶ m from the surface, will be

PE = mgh =m GM h = 1.960 x 10'J (r + h)2

Now, using conservation of energy, Total energy at surface should be equal to total energy at height h,

i.e 1.960 * 10⁷ +KE_h = 5*10⁷

Thus, Kinetic energy at height h = KE_h = 5*10⁷ - 1.960 * 10⁷ =3.04* 10⁷ J

b) At maximum height , the space probe comes momentarily at rest and hence, its kinetic energy becomes zero.

So, total energy at height r+H= 8 * 10⁶ m from the center, i.e. H=5*10⁷ m from the surface, will be

$E=KE +PE = 0 + mgh=m\frac{GM}{(r+H)^{2}}H =2.450 \times 10^{7}J$

For conservation of energy to hold true, the intital kinetic energy (That would be the total energy at surface, since PE at surface =0) must be equal to the potential energy at max. height ( which is the total energy at max height, because KE is zero there).

i.e. Initial KE = 2.45 * 10⁷ J