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Suppose I am on the moon. So I am throwing a rock upwards in a velocity of $v$ with an angle of $a$ with the ground. Now we can divide the velocity in two components. One is perpendicular to the ground ($v1$) and another parallel to the ground ($v2$). Now there is gravity of moon, so the rock will gain a maximum height due to $v1$ and after that it will ascend no more.

But as there is no air resistance on moon, so there is no force to resist $v2$. So my question is 'Will the rock reach a height and continue to move parallel to the ground in a velocity of $v2$ or not?' If it won't happen, then why won't it happen?

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  • $\begingroup$ On the Moon the trajectory of the rock would be a parabola as it would be on Earth with no air resistance but with a different value of gravitational field strength. $\endgroup$ – Farcher Apr 7 '18 at 5:56
  • $\begingroup$ @Farcher Can you please give me an answer with explanation. It will be really helpful. $\endgroup$ – Theoretical Apr 7 '18 at 5:58
  • $\begingroup$ The rock will go up and then come down and eventually hit the ground. $\endgroup$ – Farcher Apr 7 '18 at 6:00
  • $\begingroup$ @Farcher then what will happen to the parallel component of $v$. Nothing is there to resist it. $\endgroup$ – Theoretical Apr 7 '18 at 6:01
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    $\begingroup$ The horizontal motion is constant but it will stop when the rock hits the ground because then the ground will exert a force on the rock. $\endgroup$ – Farcher Apr 7 '18 at 6:56
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It will be the same as on the earth(parabolic) but the value of acceleration due to gravity will be different on the moon . Basic idea: You want to know what happens to the horizontal component of velocity . Well if we talk about ideal cases , i.e. velocity is much smaller as compared to the escape velocity of moon in that case the component of gravity will bring the ball down and there might be friction acting on the ball which will reduce the velocity . If velocity is equal to escape velocity of moon then the trajectory would be a parabola and if velocity is greater than escape velocity of moon then trajectory would be most probably a hyperbola.

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  • $\begingroup$ Roy Why will the horizontal component stop, because there is nothing acting on it to stop.I am not being stubborn but I am doing this because the matter is not clear to me. $\endgroup$ – Theoretical Apr 7 '18 at 6:50
  • $\begingroup$ Maybe the ball has come down and friction is acting between moon's surface and object. $\endgroup$ – user190931 Apr 7 '18 at 17:53
  • $\begingroup$ Let's talk about gravity free space in that case the parallel component of velocity remains unchanged because due to absence of gravity the body will not come down and no friction might act between surfaces. This is just an assumption. $\endgroup$ – user190931 Apr 7 '18 at 17:55
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There is a force in the vertically downwards direction due to gravity. This is going to accelerate the rock. Now acceleration means a change in velocity, so when the rock is ascending the velocity in vertical direction would decrease, becomes zero and then finally increases in the downward direction. It is true that the horizontal component of velocity will remain constant, but as soon as the vertical height reaches zero (as the rock is coming down with its velocity increasing),the surface of moon will exert a contact force on the rock, which can either stop the rock (thus changing it's horizontal and vertical velocity to zero) or the rock can bounce off the surface due to contact force of moon.

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In order to solve the problem you make a number of assumptions.

(1) You assume that the surface of the moon is flat rather than curved.

(2) You assume that the rotation of the moon can be ignored.

Both assumptions are reasonable for throwing a rock. However if you were shooting long range artillery then the assumptions wouldn't hold.

Now with the given assumptions you split the velocity into horizontal and vertical components. Gravity interacts with the vertical component only.

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  • $\begingroup$ If gravity interacts with the vertical component only, then why does the vertical component stop. No force is being acted on it. $\endgroup$ – Theoretical Apr 7 '18 at 6:53
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The answer is yes. The parallel component remains unchanged.

A force tries to cause an acceleration in the same direction as it pulls. This is Newton's 2nd law. Any other direction in which no forces act experiences no acceleration and thus constant speed.

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  • $\begingroup$ Is this principle also used for satellites? $\endgroup$ – Theoretical Apr 7 '18 at 11:22
  • $\begingroup$ @AsifIqubal Of course. Why wouldn't it? Newton's laws are universal - they apply in space just as much as anywhere else. $\endgroup$ – Steeven Apr 7 '18 at 14:31
  • $\begingroup$ Then why don't satellites go up and come down like the rock instead of circling the earth. $\endgroup$ – Theoretical Apr 7 '18 at 15:46
  • $\begingroup$ How can I calculate how much distance the rock will cover horizontally? $\endgroup$ – Theoretical Apr 7 '18 at 15:50

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