Some deep space cosmonauts land on a small planet.  They know that the mass of their craft is 500 tonnes and that to stop, they had to use 6.0 × 10¹¹ J of energy.  They also worked out that the radius of the planet is 1500 km. 

(a)     Explain why the potential energy is - 6.0 × 10¹¹ J.  Why is the sign negative?

(b)    What is the mass of the planet? 

(c)     What is its density?

(a) If they had to use 6.0 × 10¹¹ J to stop, then they would have gained 6.0 × 10¹¹ J as kinetic energy. (P)

This would have come from the potential energy, which would be - 6.0 × 10¹¹ J (P)

The minus sign indicates that work has been got out while approaching the planet. (P)

(b) Conversions first: 500 tonnes = 500 × 10³ kg, radius = 1.5 × 10⁶ m (P)

Formula:  E_p = -GMm

                              r

Rearranging: M = -E_pr

                               Gm (P)

M = - - 6.0 × 10¹¹ J × 1.5 × 10⁶ m                      (P)

             6.67 × 10^-11 Nm² kg^-2 × 500 × 10³  kg

M = 2.70 × 10²² kg (P)

(c) Volume = 4/3 pr³ = 4/3 × p × (1.5 × 10⁶ m)³ = 1.41.× 10¹⁹ m³ (P)

Density = mass ÷ volume = 2.70 × 10²² kg ÷ 1.41.× 10¹⁹ m³ (P)

Density =  1900 kg/m³ (P)