Gravity problems - when is the Earth's attraction 5 Newtons

[Sedm]

I just recently started studying physics about a month ago, so I'm not too good at it yet. Anyway, here's the problem:

When a certain object is 100 meters above the surface of the Earth, Earth's attraction for it is 10 Newtons. In order for Earth's attraction for the same object to be only 5 Newtons, the object must be taken to a distance from the surface of Earth of _______.

My approach to the problem: I thought that in order to decrease the force by half, you would have to quadruple the distance (in other words, it varies inversely as the distance between the two points squared). This led me to think that the answer is 400 meters. Naturally, as I don't know that much about physics yet though, I don't think I'm right.

Help, anyone?

 

[zwtipp05]

Your reasoning is close. The Universal Law of Graviation is [tex]G\frac{m_1m_2}{r^2}[/tex]

r is the distance from the center of earth.

So to double the force of gravity, you would increase r by a factor of [tex]\sqrt{2}[/tex]

 

[HallsofIvy]

I just recently started studying physics about a month ago, so I'm not too good at it yet. Anyway, here's the problem:

When a certain object is 100 meters above the surface of the Earth, Earth's attraction for it is 10 Newtons. In order for Earth's attraction for the same object to be only 5 Newtons, the object must be taken to a distance from the surface of Earth of _______.

My approach to the problem: I thought that in order to decrease the force by half, you would have to quadruple the distance (in other words, it varies inversely as the distance between the two points squared). This led me to think that the answer is 400 meters. Naturally, as I don't know that much about physics yet though, I don't think I'm right.

Help, anyone?

You have that exactly backwards! Since force is inversely proportional to distance squared, if you double the distance, the force is cut by 1/4.

 

[mukundpa]

1.
To halved the force the distance is to be [tex]\sqrt{2} [/tex] times the origional distance from the center of earth.
2.
100 m is quite small as compared to the radius of Earth which is nearly 6380000 m and hence the body is still supposed at the surface of the earth.
3.
The new distance is [tex]\sqrt{2} [/tex] times 6380000 m from the center of earth.