The Ingenuity Mars helicopter (above) has a mass of 1.8kg. and climbs to a height of 75m. This decreases the chemical store in its battery by 500J. What is the gravitational field strength g on Mars?

Answer:

3.7 m/s^2

Explanation:

Lets start by calculating the gain in potential energy when the Ingenuity rises to 75 meters from Mars' surface.  It has a mass of 1.8kg.  The potential energy is given by PE = mgh, where m is mass, g is the acceleration due to Mars' gravity, and h is height, in meters.

PE = (1.8kg)(75 meters)(g)

PE = (135 kg*m)*g

This required 500J of energy from the battery.  1 Joule = 1 kg*m^2/s^2

That means the PE gain is equal to 500 kg*m^2/s^2.

Set this equal to the PE we first calculated:

 500 kg*m^2/s^2 = (135 kg*m)*g

g = (500 kg*m^2/s^2)/(135 kg*m)

g = 3.7 m/s^2

The published value is 3.721 m/s^2

Answer:

Approximately [tex]3.7\; {\rm N\cdot kg^{-1}}[/tex] assuming that the energy from the battery was entirely converted into gravitational potential energy.

Explanation:

Near the surface of a planet, the gravitational field is approximately uniform. Let [tex]g[/tex] denote the gravitational field strength. When an object of mass [tex]m[/tex] is lifted up and its height increases by [tex]\Delta h[/tex], the gravitational potential energy (GPE) of this object would have increased by [tex]m\, g\, \Delta h[/tex].

Under the assumptions, the gravitational potential energy of this helicopter would have increased by [tex]500\; {\rm J}[/tex]. In other words:

[tex]\begin{aligned}& m\, g\, \Delta h = (\text{change in GPE}) = 500\; {\rm J}\end{aligned}[/tex].

It is given that [tex]m = 1.8\; {\rm kg}[/tex] and [tex]\Delta h = 75\; {\rm m}[/tex]. Rearrange the equation above to find gravitational field strength [tex]g[/tex]:

[tex]\begin{aligned}g &= \frac{(\text{change in GPE})}{m\, \Delta h} \\ &= \frac{(500\; {\rm J}))}{(1.8\; {\rm kg})\, (75\; {\rm m})} \approx 3.7\; {\rm N\cdot kg^{-1}}\end{aligned}[/tex].

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