gravity and force stuff

[pizza1994]

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If you drop a ball on the surface of the Earth it will accelerate to the ground at 10 m/s2 (neglecting air resistance). What would be the acceleration of a ball twice the mass if dropped from the same height on a planet that is 10 times more massive than Earth?

answer is 100 m/s^2

I got 20 m/s^2 using the GMm/(r^2) equation where m= 2m and M=10 and therefore gravitaional force is 20 m/s^2.....

Can someone help we on this question as well please?

 

[DrknoSDN]

The mass of the ball is a distractor. Knowing on earth all masses accelerate at the same rate you can ignore the mass of the ball.
From that point if you increase the mass of earth by 10, the acceleration will increase by 10 also.

http://www.physicsclassroom.com/class/circles/Lesson-3/The-Value-of-g
Bottom of page, Calculating g on other planets.
Equation is g = GM/r^2 (ignores mass of falling object)

 

[pizza1994]

The mass of the ball is a distractor. Knowing on earth all masses accelerate at the same rate you can ignore the mass of the ball.
From that point if you increase the mass of earth by 10, the acceleration will increase by 10 also.

http://www.physicsclassroom.com/class/circles/Lesson-3/The-Value-of-g
Bottom of page, Calculating g on other planets.
Equation is g = GM/r^2 (ignores mass of falling object)

so then in what cases do we use the g= GM1m2/r^2 equation. Cause for the equation you stated I see that for that to work then you assume your object of interest is on earth and hence has weight=mg

 

[orangetea]

so then in what cases do we use the g= GM1m2/r^2 equation. Cause for the equation you stated I see that for that to work then you assume your object of interest is on earth and hence has weight=mg

From the problem I've done that equation is useful when they give you problems involving the earth and some other object in space. It gives you the acceleration felt by an object at given altitudes. For objects on earth just stick to the method above.

 

[pizza1994]

From the problem I've done that equation is useful when they give you problems involving the earth and some other object in space. It gives you the acceleration felt by an object at given altitudes. For objects on earth just stick to the method above.

Thanks so much! 🙂

 

[pizza1994]

The mass of the ball is a distractor. Knowing on earth all masses accelerate at the same rate you can ignore the mass of the ball.
From that point if you increase the mass of earth by 10, the acceleration will increase by 10 also.

http://www.physicsclassroom.com/class/circles/Lesson-3/The-Value-of-g
Bottom of page, Calculating g on other planets.
Equation is g = GM/r^2 (ignores mass of falling object)

k sorry this is from another question I posted but for the following question: A 0.2-kg apple rests on the surface of the Earth. Approximate the gravitational force exerted by the apple on the Earth. (Note: The universal gravitational constant, G, is 6.7 × 10-11 N-m2/kg2, the mass of the Earth is 6.0 × 1024 kg, and the radius of the Earth is 6.4 × 106 m.)ANSWER= 2 N

why in this question is it that I can't ignore the mass of the apple since all masses accelerate at the same rate on earth. and so I would just use g=GM(earth) / (r^2) and this doesnt give me 2! I dont get this anymore

 

[DrknoSDN]

Gravity is going to accelerate all objects at the same "rate" of about 10m/s^2.

If you have a 1kg and a 2kg ball, they fall at the same speed. However the force exerted on the 2kg ball is going to need to be double because it requires more force to accelerate something (F=ma) where mass is constant.

A 1kg object always experiences a 10N force of gravity near the surface of earth. It will always always fall with the same acceleration on earth. The force is dependent on the mass but the acceleration is not.
For the equation F=ma, just remember that a = g = GM/r^2 . ( a constant for each planet, but each planet has a different value based on the planets mass and radius )