Excel in math and science.

Join using Facebook Join using Google Join using email

Forgot password? New user? Sign up

Existing user? Log in

Your answer seems reasonable. Find out if you're right!

That seems reasonable. Find out if you're right!

Join using Facebook Join using Google Join using email

Forgot password? New user? Sign up

Existing user? Log in

Classical Mechanics

Newton's Law of Gravity

Concept Quizzes

Newton's Laws of Gravity
Gravitational Potential Energy
Newton's Law of Gravity - Problem Solving

Challenge Quizzes

Newton's Law of Gravity: Level 2-4 Challenges

Gravitational Potential Energy

What is the approximate gravitational potential energy of the two-particle system, of masses $5.4\text{ kg}$ and $2.9 \text{ kg},$ separated by a distance of $15.0\text{ m}?$

Assumptions and details

The universal gravitational constant is $G=6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2\text{/kg}^2.$

-4.9 \times 10^{-11} \text{ J}

-7.0 \times 10^{-10} \text{ J}

-7.0 \times 10^{-11} \text{ J}

-4.9 \times 10^{-10} \text{ J}

Show explanation

View wiki

by Brilliant Staff

Consider an $xy$ -plane in deep space, where two particles A and B are located. Particle A is fixed at the origin and particle B is apart from particle A with a distance of $3.6\text{ m}.$ The masses of particles A and B are $24\text{ kg}$ and $12\text{ kg},$ respectively. If particle B is released from rest, what is the kinetic energy of B when it has moved $0.1\text{ m}$ toward A?

Assumptions and Details

The universal gravitational constant is $G=6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2\text{/kg}^2.$

1.5 \times 10^{-9}\text{ J}

3.0 \times 10^{-10}\text{ J}

1.5 \times 10^{-10}\text{ J}

3.0 \times 10^{-9}\text{ J}

Show explanation

View wiki

by Brilliant Staff

The mean diameter of Mars is $6.7\times 10^3\text{ km}$ and its mass is $6.7\times 10^{23}\text{ kg}.$ What is the approximate escape speed on Mars?

Assumptions and Details

The universal gravitational constant is $G=6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2\text{/kg}^2.$

5.2\text{ km/s}

1.6\text{ km/s}

2.3\text{ km/s}

2.8\text{ km/s}

Show explanation

View wiki

by Brilliant Staff

Four identical particles are forming a square with side length $12.0\text{ m},$ as shown in the above figure. Then the side length of the square is reduced to $4.0\text{ m}.$ If the mass of each particles is $19.0\text{ kg},$ approximately how much gravitational potential energy of the four-particle system is reduced, assuming that the four particles are point particles?

Assumptions and Details

The universal gravitational constant is $G=6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2\text{/kg}^2 ?$

-2.2 \times 10^{-8}\text{ J}

-2.2 \times 10^{-6}\text{ J}

-3.3 \times 10^{-7}\text{ J}

-3.3 \times 10^{-5}\text{ J}

Show explanation

View wiki

by Brilliant Staff

If a $6000 \text{ kg}$ space rocket is launched vertically from the surface of the Earth with initial energy $7.0 \times 10^{11} \text{ J},$ what will be the kinetic energy of the space rocket when it is $1.4 \times 10^7 \text{ m}$ from the center of the Earth, assuming that the mass and the radius of the Earth are $6.0 \times 10^{24}\text{ kg}$ and $6.4 \times 10^6 \text{ m},$ respectively?

Assumptions and Details

The universal gravitational constant is $G=6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2\text{/kg}^2.$

2.9 \times 10^{11}\text{ J}

5.0 \times 10^{11}\text{ J}

4.4 \times 10^{11}\text{ J}

3.5 \times 10^{11}\text{ J}

Show explanation

View wiki

by Brilliant Staff