At point X shown in the figure, which is midway between the centers of two isolated planets, the net gravitational force on an object is zero. If the mass of the smaller planet is M, the mass of the larger planet is 3. (A) M/4 (B) M/2 (C) M (D) 2M
Added by Michelle C.
Close
Step 1
This means that the gravitational force of the smaller planet on the object is equal in magnitude but opposite in direction to the gravitational force of the larger planet on the object. Show more…
Show all steps
Your feedback will help us improve your experience
Karan Soni and 63 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Ivan K.
Assume that $L$ is much larger than the radius of either planet. What is the position, $x$, of the spacecraft (given as a function of $L, M_{1}$, and $M_{2}$ ) if the net force on the spacecraft is zero?A spaceship of mass $m$ is located between two planets of masses $M_{1}$ and $M_{2}$; the distance between the two planets is $L$, as shown in the figure.
In Fig. $14-18 b$, two spheres of mass $m$ and a third sphere of mass $M$ form an equilateral triangle, and a fourth sphere of mass $m_{B}$ is at the center of the triangle. The net gravitational force on that central sphere from the three other spheres is zero. (a) What is $M$ in terms of $m ?$ (b) If we double the value of $m_{B}$, what then is the magnitude of the net gravitational force on the central sphere?
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD