Calculating Force of Attraction between Earth and 3.0 kg Mass

[sheri1987]

Homework Statement

Calculate, using Newton's law of gravity, the size of the force of attraction between the Earth and a mass of 3.0 kg on the Earth. Data: Distance to the center of Earth from the surface = 6370 km. Mass of Earth = 5.98 1024 kg. Gravitational constant (G = 6.67 times 10-11 Nm2/kg2).

Homework Equations

F = G(m1 * m2)/ r^2

The Attempt at a Solution

I plugged the numbers into the equation and converted km to meters and kg to g ...

6.67E-11 * (3000)(5.98E27) / (6370000)^2
and I got 2.949E-7, but it is wrong I think

 

[neutrino]

The answer is quite a way off.

1.Why did you convert Kg to g? The SI unit of mass is Kg. When you substitute 6.67e-11 for G, it means G = 6.67e-11 Nm²/Kg². So make sure the units are uniform throughout.

2.You can make a rough estimate of the answer. If you've come across the term acceleration due to Earth's gravity(g), you may know that it is roughly equal to 10ms^-2. The weight of an object on the Earth, which is nothing but the gravitational force due to the Earth, is mg, where m is the mass of the object.

 

[m2003]

I would like to point out that your attempt at a solution is close, but there are a few errors in your calculations. First, the mass of the Earth should be 5.98E24 kg, not 5.98E27 kg. Additionally, when converting from kilometers to meters, you should multiply by 1000, not 100. Finally, when converting from kilograms to grams, you should divide by 1000, not multiply. Correcting these errors, the calculation should look like this:

F = (6.67E-11 Nm^2/kg^2) * ((3.0 kg) * (5.98E24 kg)) / (6370000 m)^2
= 2.949E-7 N

Therefore, the force of attraction between the Earth and a mass of 3.0 kg on the Earth is approximately 2.949E-7 Newtons.