Practice with proportions: Two masses a distance R apart attract each other with

Question

Practice with proportions:
Two masses a distance R apart attract each other with a gravitational force F. For each of the cases below, give the factor by which F is changed, if:
a.) if R is multiplied by 2, then Fnew = ?· Fold
b.) if Ris multiplied by 5.2, then Fnew = ? · Fold
c.) if R is divided by 4, then Fnew = ? · Fold


Mass A (11 kg) and mass B (4 kg) sit on the ground, 2 m apart.

a.) What is size of the the gravitational force:
on A from B? N
on B from A? N
b.) What is the size of the gravitational force:
on A from the Earth? N
on B from the Earth? N

c.) Explain why we normally only worry about the gravitational force on such objects from the Earth, and don't worry about including the gravitational force on objects from each other. ?
d.) Do A and B exert forces back on the Earth itself? If so, are those forces of smaller, equal, or larger size than your answers for part b?
smaller
same size
larger

Explanation / Answer

The distance between the masses = R The gravitational force betweem the masses = F And the force is given by the formula, F = GMm/R^2 a) If the R is multiplied by 2, then the new force is              Fnew = GMm/(2R)^2 = GMm/4R^2 = (1/4)GMm/R^2 = (1/4)F So the new force is (1/4) times of the old force b) If the R is multiplied by 5.2, then the new force is        Fnew = GMm/(5.2R)^2 = GMm/27.04R^2 = (1/27.04)GMm/R^2 = (1/27.04)F So the new force is (1/27.04) times of the old force c)  If the R is devided by 4, then the new force is              Fnew = GMm/(R/4)^2 = 16*GMm / R^2 = 16GMm/R^2 = 16F So the new force is 16 times of the old force PLEASE POST THE QUESTIONS SEPERATELY. THANq        Fnew = GMm/(5.2R)^2 = GMm/27.04R^2 = (1/27.04)GMm/R^2 = (1/27.04)F So the new force is (1/27.04) times of the old force c)  If the R is devided by 4, then the new force is              Fnew = GMm/(R/4)^2 = 16*GMm / R^2 = 16GMm/R^2 = 16F So the new force is 16 times of the old force PLEASE POST THE QUESTIONS SEPERATELY. THANq              Fnew = GMm/(R/4)^2 = 16*GMm / R^2 = 16GMm/R^2 = 16F So the new force is 16 times of the old force PLEASE POST THE QUESTIONS SEPERATELY. THANq

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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