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

In summary, when a .36 kg apple is driven, it experiences a force from the Earth that accelerates it at 9.8 m/s squared towards the Earth's surface. By Newton's third law, the apple also exerts an equal but opposite force on the Earth. With the mass of the Earth given as 5.89x10^24 kg, the magnitude of the Earth's acceleration towards the apple can be calculated using the expression .36*9.8/5.98x10^24.
  • #1
KellySierra22
1
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
  • #2


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

[itex] F_{apple \rightarrow earth}= m_{apple} \times g_{apple \rightarrow earth} [/itex]

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

[itex] F_{apple \rightarrow earth}=F_{earth \rightarrow apple}[/itex]

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

[itex] F_{earth \rightarrow apple}= m_{earth} \times g_{earth \rightarrow apple} [/itex]

You must solve for the acceleration of the Earth towards the apple.
 
  • #3


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?
 

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

1. How does Earth exert a force on a .36 kg apple when it is being driven?

Earth exerts a force on any object due to its gravitational pull. The force is directly proportional to the mass of the object and inversely proportional to the distance between the object and Earth's center.

2. Why does the mass of the apple matter in this scenario?

The mass of an object determines the amount of gravitational force it experiences. In this case, the .36 kg apple has a smaller mass compared to Earth, so it experiences a smaller force.

3. How does the distance between the apple and Earth affect the force?

The force of gravity decreases as the distance between two objects increases. So, the farther the apple is from Earth, the weaker the force of gravity it experiences.

4. Is the force exerted by Earth the only force acting on the apple?

No, there are other forces at play, such as friction and air resistance. However, in this scenario, we are only considering the force of gravity exerted by Earth on the apple.

5. Does Earth experience a force from the apple as well?

Yes, according to Newton's third law of motion, for every action, there is an equal and opposite reaction. So, while Earth is exerting a force on the apple, the apple is also exerting a force on Earth, but it is much smaller due to Earth's larger mass.

Back
Top