When you drive a .36 kg apple, Earth exerts a force on It that

AI Thread Summary
When a .36 kg apple is driven downward, Earth exerts a force that accelerates it at 9.8 m/s². According to Newton's third law, the apple exerts an equal and opposite force on Earth. To find Earth's acceleration towards the apple, one must apply the formula F=ma, using the apple's mass and the gravitational force it experiences. The discussion emphasizes the importance of correctly identifying forces and masses in calculations. Ultimately, the numerical solution for Earth's acceleration is derived from the relationship established by these forces.
KellySierra22
Messages
1
Reaction score
0
When you drive a .36 kg apple, Earth exerts a force on It that accelerates it at 9.8 m/s squared toward the Earth's surface. According to Newtons third law the Apple must exert an equal but opposite force on earth. If the mass of the Earth is 5.89x10^24 kg,what is the magnitude of the Earth's acceleration toward the apple?

I have tried this problem a couple of times and get it wrong. My math is: .36*9.8/5.98x10^24.
 
Physics news on Phys.org


Hello KellySierra22,
Instead of giving your workings with numbers, it is easier to understand the problem in algebra.

We start with the well known F=ma, however we must be careful about which force and which mass (although the acceleration is constant). So I label them

F_{apple \rightarrow earth}= m_{apple} \times g_{apple \rightarrow earth}

That is the force the Earth exerts on the apple is the mass of the apple times acceleration, you must calculate this force first. Now that you have done so we can use Newtons third law

F_{apple \rightarrow earth}=F_{earth \rightarrow apple}

It may help to draw this pictorially. So you now have an equation relating the force you already have to the force the apple exerts on the earth. You can now calculate the acceleration on the Earth from the apple using

F_{earth \rightarrow apple}= m_{earth} \times g_{earth \rightarrow apple}

You must solve for the acceleration of the Earth towards the apple.
 


KellySierra22 said:
When you drive a .36 kg apple, Earth exerts a force on It that accelerates it at 9.8 m/s squared toward the Earth's surface. According to Newtons third law the Apple must exert an equal but opposite force on earth. If the mass of the Earth is 5.89x10^24 kg,what is the magnitude of the Earth's acceleration toward the apple?

I have tried this problem a couple of times and get it wrong. My math is: .36*9.8/5.98x10^24.

Hello KellySierra22, Welcome to Physics Forums.

The expression that you have shown should give you the numerical value for the acceleration of the Earth towards the apple. What value did you obtain when you completed the math?
 
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...
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 .
Back
Top