Gravitational force is given by, ![F= G\frac{mM}{R^{2}}](https://tex.z-dn.net/?f=%20F%3D%20G%5Cfrac%7BmM%7D%7BR%5E%7B2%7D%7D%20%20%20)
Where, m and M are the masses of the objects, R is the distance between them and G gravitational constant.
Gravitational force of the star on planet 1, ![F_{1}= G\frac{m_{1}M}{R^{2}}](https://tex.z-dn.net/?f=%20F_%7B1%7D%3D%20G%5Cfrac%7Bm_%7B1%7DM%7D%7BR%5E%7B2%7D%7D%20%20%20)
Gravitational force of the star on planet 2, ![F_{2}= G\frac{3m_{1}M}{(3R)^{2}}](https://tex.z-dn.net/?f=%20F_%7B2%7D%3D%20G%5Cfrac%7B3m_%7B1%7DM%7D%7B%283R%29%5E%7B2%7D%7D%20%20%20%20%20)
Ratio, ![\frac{F_{1}}{F_{2}}= \frac{\frac{Gm_{1}M}{R^{2}}}{\frac{G3m_{1}M}{(3R)^{2}}}](https://tex.z-dn.net/?f=%20%5Cfrac%7BF_%7B1%7D%7D%7BF_%7B2%7D%7D%3D%20%5Cfrac%7B%5Cfrac%7BGm_%7B1%7DM%7D%7BR%5E%7B2%7D%7D%7D%7B%5Cfrac%7BG3m_%7B1%7DM%7D%7B%283R%29%5E%7B2%7D%7D%7D%20%20%20%20%20%20%20%20)
![\frac{F_{1}}{F_{2}}= \frac{3}{1}](https://tex.z-dn.net/?f=%20%5Cfrac%7BF_%7B1%7D%7D%7BF_%7B2%7D%7D%3D%20%20%5Cfrac%7B3%7D%7B1%7D%20%20%20%20)
Therefore, the gravitational force of the star on the planet 1 is three times that on planet 2.
Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
Usually, it increases the solubility in water.