Question

Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an 80.0-kg astronaut in orbit moving at 27,500 km/h.

Short answer

The kinetic energy of the astronaut is \(250\) time of the moving truck.

Step-by-step solution

Definition of Concept

Kinetic energy: The energy stored in the body by the virtue of its motion is called kinetic energy.

Let us consider a body of mass\(m\)moving with velocity\(v\), then the kinetic energy associated with the body is,

\(KE = \frac{1}{2}m{v^2}\)

Calculate kinetic energy of the truck

The kinetic energy of the truck is,

\({\rm{K}}{{\rm{E}}_T} = \frac{1}{2}{m_T}v_T^2\)

Putting all known values,

\(\begin{aligned}K{E_T} &= \frac{1}{2} \times \left( {20000{\rm{ kg}}} \right) \times {\left( {110{\rm{ km}}/{\rm{h}}} \right)^2}\\ &= \frac{1}{2} \times \left( {20000{\rm{ kg}}} \right) \times {\left( {\left( {110{\rm{ km}}/{\rm{h}}} \right) \times \left( {\frac{{1000{\rm{ m}}}}{{1{\rm{ km}}}}} \right) \times \left( {\frac{{1{\rm{ h}}}}{{3600{\rm{ s}}}}} \right)} \right)^2}\\ \approx 9.34 \times {10^6}{\rm{ J}}\end{aligned}\)

Therefore, the required kinetic energy of the truck is \(9.34 \times {10^6}{\rm{ J}}\).

Calculate kinetic energy of the astronaut

The kinetic energy of astronaut is,

\(K{E_A} = \frac{1}{2}{m_A}v_A^2\)

Putting all known values,

\(\begin{aligned}K{E_A} &= \frac{1}{2} \times \left( {80{\rm{ kg}}} \right) \times {\left( {27500{\rm{ km}}/{\rm{h}}} \right)^2}\\ &= \frac{1}{2} \times \left( {80{\rm{ kg}}} \right) \times {\left( {\left( {27500{\rm{ km}}/{\rm{h}}} \right) \times \left( {\frac{{1000{\rm{ m}}}}{{1{\rm{ km}}}}} \right) \times \left( {\frac{{1{\rm{ h}}}}{{3600{\rm{ s}}}}} \right)} \right)^2}\\ &= 2.33 \times {10^9}{\rm{ J}}\end{aligned}\)

Therefore, the required kinetic energy of the astronaut is \(2.33 \times {10^9}{\rm{ J}}\).

Comparing the kinetic energy of the truck and the astronaut

The ratio of kinetic energy of the truck and the astronaut is,

\(\begin{aligned}\frac{{K{E_A}}}{{K{E_T}}} = \frac{{2.33 \times {{10}^9}{\rm{ J}}}}{{9.34 \times {{10}^6}{\rm{ J}}}}\\\frac{{K{E_A}}}{{K{E_T}}} \approx 250\\K{E_A} \approx 250K{E_T}\end{aligned}\)

Therefore, the kinetic energy of the astronaut is \(250\) time of the moving truck.