Quiz: Calculating Gravitational Forces and Orbital Speeds

[PhysWiz3.14]

I have to make a sample quiz for class. Can someone look these over and answer them so I can make sure my work is correct

1. The value of g at the Earth's surface is about 10 m/s2. What is its value at a distance from the Earth's center that is three times the Earth's radius?

2.Calculate the speed in meters per second at which the Earth revolves around the Sun. Assume the orbit is nearly circular, with radius r=1.5 10^11 m.

3. Somewhere between the Earth and the Moon, the gravitational pull exerted by these two bodies on a space pod would cancel. Is this location nearer the Earth or nearer the Moon?
A) Nearer the Earth.
B) Nearer the Moon.
C) At the half-way point.

Thanks for the help in advance.

 

[Dick]

Help in advance requires some work in advance. Nothing is free. Pick one of those problems and tell us what your answer is and why you think so and you will get any number of responses.

 

[Moetasim]

1. The value of g at a distance of three times the Earth's radius would be 2.22 m/s2. This can be calculated using the formula g = Gm/R^2, where G is the gravitational constant, m is the mass of the Earth, and R is the distance from the Earth's center. Plugging in the values, we get g = (6.67x10^-11 Nm^2/kg^2)(5.97x10^24 kg)/(3x(6.38x10^6 m))^2 = 2.22 m/s2.

2. The speed at which the Earth revolves around the Sun can be calculated using the formula v = √(GM/r), where G is the gravitational constant, M is the mass of the Sun, and r is the distance between the Earth and the Sun. Plugging in the values, we get v = √((6.67x10^-11 Nm^2/kg^2)(1.99x10^30 kg)/(1.5x10^11 m)) = 29,800 m/s.

3. The location where the gravitational pull of the Earth and Moon cancel out, known as the Lagrange point, is located closer to the Moon. This point is approximately 60% of the distance from the Earth to the Moon. Therefore, the correct answer is B) Nearer the Moon.