Universal Gravitation Problem (Grade 11)

AI Thread Summary
To determine how much higher an object would reach when thrown from the moon compared to Earth, the gravitational acceleration values of both celestial bodies are crucial, with the moon's g at 1.6 N/kg and Earth's at 9.81 N/kg. The ratio of these accelerations suggests that the object would reach approximately 6.13 times higher on the moon. While the initial calculations seem straightforward, incorporating kinematic equations, such as v^2 = 2*a*h, can provide a more rigorous analysis. Considering the gravitational force equation Fg = (G*m1*m2)/d^2 may not be necessary for this problem, as g already encapsulates those variables. Ultimately, the initial impulse is likely insufficient to reach altitudes where gravitational variation would significantly impact the results.
danielle_54321
Messages
2
Reaction score
0

Homework Statement



If you threw something vertically upward from the moon's surface where g=1.6N/kg [down], how many times higher would it reach compared to throwing it from the surface of the Earth with the same force? Neglect the effect of air resistance on the Earth.

Homework Equations


I don't know which equations to use. Without other variables, I don't know where to start. I am not necessarily looking for a solution, I just need help at getting my solution started.
I do know that
g=9.81N/kg on earth.
F=ma is likely a relevant equation, becuase g is an acceleration and the force and mass are both constants.

The Attempt at a Solution



g(earth)/ gmoon
=9.81/1.6
=6.13125

Does this make sense? The solution seems much too simple. :(
 
Physics news on Phys.org
well, sometimes one gets lucky. for more rigor, you might add kinematics expression relating distance and velocity, such as v^2=2*a*h where v is the initial velocity the rock or whatever was thrown upwards.
 
Okay, thank-you.

So, if I do not have any other varables (height, acceleration, force, time, distance, initial velocity or final velocity) is my answer the most accurate possible?

Also, do you think I should factor in:

Fg = (G*m1*m2)/d^2

where G is the gravitational constant (6.67 x 10^-11) and m1 is the mass of the planet and m2 is the mass of the object thrown and d is the distance between the mass' centre and the centre of the object?


In order to do any further calculations, would it profit me to determine the mass and radius of both the moon and the earth?

I don't know if that makes any sense or not?
 
g already is G*m1/d^2. An it is fair to assume that the initial impulse is not powerful enogh to reach an altitude for which you need to consider any variation on d.

If you want, you can calculate the highest point for both situations. You will get that result again. Or using the formula by denverdoc, as you have the same initial speed, it is easy to see that 'a' and 'h' wil be inversely related.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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...
Back
Top