Mars Surface Curvature: 2m/3600m

  • Thread starter Thread starter touqra
  • Start date Start date
AI Thread Summary
The curvature of Mars results in a vertical drop of 2 meters over a horizontal distance of 3600 meters. This means that if a straight edge were laid on the Martian surface, it would be 2 meters above the surface after traveling 3600 meters. Consequently, for a person who is 2 meters tall, the horizon on Mars would only be 3600 meters away. This curvature is significant for understanding visibility and distance on Mars. The discussion emphasizes the implications of this curvature for exploration and observation on the planet.
touqra
Messages
284
Reaction score
0
What does it mean by the sentence:

The curvature of Mars is such that its surface drops a vertical distance of 2 meters for every 3600 meters tangent to the surface.
 
Physics news on Phys.org
A tangent to a circle is a straight line that just about touches the edge of the circle somewhere (anywhere). So imagine a huge straight edge resting on Mars' surface. If you were to follow that straight edge from the point it touches Mars for 3600 meters, you would find that the straight edge is now 2 meters above the surface of Mars, such is the curvature of Mars.
 
See diagram A for a literal translation of the sentence.
See diagram B for its significance.

It means that, to a 2m tall person, Mars' horizon is a mere 3600m away.
 
Last edited:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top