For the last question, I stared at it for so long since there was something that just seemed off to me. First off, I think, if we disregard the direction of g (i.e. making the g graph curve on the same side of the x-axis), we will lose marks. Because g is always a vector; the gravitational force formula is F = GMm/R^2 and the 2nd law formula is F = ma, from this we can conclude that the acceleration here is GM/R^2, so it does have a direction and it is a must. Also, the value of potential gradients at surface of either side of the objects seems wrong too, I second the part where "g value must be divided by 3 since the mass of the moon is way smaller than that of the planet"; however, we have to be aware that the formula for g is GM/R^2 so not only M is taken into account, but also the radius R. Apparently, from the planet to moon (2R --> R) it is divided by 2. Notably, there is an inverse square relationship between the two variables so multiplying R by 1/2 makes the g value to be multiplied by 4. If we consider this factor change, the potential gradient g at surface of moon should be 4/3R. The factor 4/3 is bigger than 1 so it is not a decrease but an increase, if we set 30 small squares for the planet, we should need 40 small squares for the moon's gravitational field strength.
For the last question, I stared at it for so long since there was something that just seemed off to me.
First off, I think, if we disregard the direction of g (i.e. making the g graph curve on the same side of the x-axis), we will lose marks. Because g is always a vector; the gravitational force formula is F = GMm/R^2 and the 2nd law formula is F = ma, from this we can conclude that the acceleration here is GM/R^2, so it does have a direction and it is a must.
Also, the value of potential gradients at surface of either side of the objects seems wrong too, I second the part where "g value must be divided by 3 since the mass of the moon is way smaller than that of the planet"; however, we have to be aware that the formula for g is GM/R^2 so not only M is taken into account, but also the radius R. Apparently, from the planet to moon (2R --> R) it is divided by 2. Notably, there is an inverse square relationship between the two variables so multiplying R by 1/2 makes the g value to be multiplied by 4. If we consider this factor change, the potential gradient g at surface of moon should be 4/3R. The factor 4/3 is bigger than 1 so it is not a decrease but an increase, if we set 30 small squares for the planet, we should need 40 small squares for the moon's gravitational field strength.
Thank you
Thank youu
Glad it helped😀
come back
Arigato
i think the graph is wrong
Yes it should be a mirror of what she drew but overall she is fantastic❤
Thank you