Suppose you are in a hot air balloon with a sandbag that has a certain mass. The hot air balloon is moving upwards at a constant velocity of $15$ $m$.$s^{-1}$. If you throw the sandbag out of the hot air balloon, will the velocity of the hot air balloon change?

I thought that it will increase, because according to Newton's first law, an object will move in uniform motion unless an unbalanced force acts upon it. If the sandbag is released, there will be an unbalanced force. Or am I wrong? Please help.


Yes. The velocity of the balloon is determined by the buoyant force of the balloon, determined by its effective density and volume balanced against the density of the surrounding air, balanced against the aerodynamic drag of the balloon, which increases with speed.

When you drop a weight, the buoyancy of the balloon increases. This will cause the upward velocity to increase until the drag on the balloon matches the new buoyancy.

This ignores the effect of decreasing atmospheric density with altitude, which will also, by limiting the buoyancy, cause the rate of climb of the balloon to decrease with altitude until a maximum altitude is reached.

  • 1
    $\begingroup$ This is basically correct, although I wouldn't have said that the buoyancy increases. Rather, the buoyant force on the balloon is roughly constant while its weight decreases, resulting in an increased net upwards force. It depends on whether you view the buoyant force as "force from surrounding fluid" or "force from surrounding fluid minus weight." $\endgroup$ – Michael Seifert Jun 2 '15 at 18:47

Firstly... I don't think a 'hot air' balloon will need sandbags because you can simply regulate the temperature of the air inside the balloon to take it up or get it down. Helium balloons will definitely need sandbags. Coming to the question... The sandbags are there to counter the buoyant force acting on the balloon... Lets consider the mass of the balloon be 'M' kg and the mass of sandbags be 'm' kg. Net force acting on the balloon will be ((volume of balloon) * (density of air) * (g)) -((M+m) *g). Once the sandbag is thrown away... Force will be ((volume of balloon) * (density of air) * (g)) -((M) *g). Thus the net upward acceleration of the balloon has increased which will bring about a change in the velocity of the balloon. The velocity will continue to increase until the viscous drag manages to balance the net resultant upward force. Hope this helps in clearing your doubts... (edit @dave coffman: hot air balloons do need sandbags to handle emergency situations like say a burner failure. In such an event... The sandbags can be dropped off so as to keep the descent rate safely low.)

  • $\begingroup$ Most "hot air" balloons need and use sandbags - temperature control is too gradual to be use in an emergency situation. $\endgroup$ – Dave Coffman Jun 2 '15 at 18:04
  • $\begingroup$ Just wondering How would you quickly come down in an emergency situation? $\endgroup$ – Pranav Jun 2 '15 at 18:07
  • $\begingroup$ Usually you don't need to crash-dive a balloon. Venting air through the top of the balloon typically can lower you fast enough. If a balloon crashes and the envelope is still somewhat buoyant and being blown around dangerously, it has some weakened spots where it can be cut open to rapidly let the air out, but this is dangerous and suicidal to do in flight. $\endgroup$ – Dave Coffman Jun 2 '15 at 18:12

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