The perihelion and aphelion distances for Mercury are 45.9 times 10^9 m and 69.8

Question

The perihelion and aphelion distances for Mercury are 45.9 times 10^9 m and 69.8 times 10^9 m respectively. The speed of Mercury at aphelion is 3.88 times 10^4 m/s. What is the speed at perihelion? Ans: 5.9 times 10^4 m/s. A meteoroid (a chunk of rock) is initially at rest in interplanetary space at a large distance from the Sun. Under the influence of gravity, the meteoroid begins to fall toward the Sun along a straight radial line. With what speed does it strike the sun? The radius of the sun is 6.96 times 10^8 m. Ans: 618 m/s A 50 kg woman and an 80 kg man stand on the two ends of a seesaw of length 3.00 m. Treating them as particles and ignoring the mass of the seesaw, find the center of mass of this system. Verify the conditions of equilibrium for the system. A uniform board of weight 40.0 N supports two children weighing w_1 = 350 N and w_2 = 500 N. If the support (fulcrum) is under the center of gravity of the board and if the 500 N child is 1.5 m (r_2) from the center, i) Determine the upward force F_p exerted on the board by the support, ii) Determine where the 350 N child should sit (length r_1) to balance the system. A merry-go-round, starting from rest attains its highest speed of v rad/s. Write down the general equation for kinetic energy for its rotational motion (including translational and rotational parts). A 50-N load is held in the hand with the forearm in the horizontal position. The biceps muscle is attached 4.00 Biceps cm from the joint and the weight is 40.0 cm from the joint. Find the upward force that the biceps exerts on the forearm and the downward force exerted by the upper arm on the forearm and acting at the joint. Neglect the weight of the forearm. Suppose you wanted to limit the force (F_R) acting on your joint to a maximum value of 670 N. (a) Under these circumstances, what maximum weight would you attempt to lift? (b) What force would your biceps apply while lifting this weight? A ball of mass M and radius R starts rolling (without slipping) from rest at a height of 2.0 m and rolls down a 30 degree slope. What is the linear speed of the ball as it leaves the incline? In case the ball is replaced by a solid cylinder of the same mass and radius as the ball and released from the same height, which one would win?

Explanation / Answer

1. Applying Kepler's law:

rp vp = ra va

(45.9 x 10^9) (vp) = (69.8 x 10^9) (3.88 x 10^4)

vp = 5.9 x 10^4 m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.