Question

FERENCES MAILINGS REVIEWVIEWFoxit PDF The force of gravity, F, exerted between two objects is equal to the product of the gravitational constant, G, the mass of the irst object, m, and the mass of the second object, ma, divided by the square of the distance between their centers, d. This is often used to determine the gravitational attraction behween two massive bodies, such as planets, in space. a. Write the formula for F described in the situation above. Type your response here: b. If you were a physicist trying to determine the gravitational constant and you were able to measure the force of gravity exerted between two objects, their masses, and the distance between their centers, how would you rewnite this formula? Type your response here: c. When the center of Earth is 2 x 10 kilometers from the center of Mars, the force of gravity between the two planets is about 64.32 x 10 newtons. The mass of Earth is about 6 x 104 kilograms, and the mass of Mars is about 6.4 x 10 kilograms. Using these values, estimate the gravitational constant Type your response here: d. Rewrite the original formula to solve for one of the mass values. Type your response here: e. When the center of Earth is 38 x 10 kilometers from the center of the moon, the force of gravity between Earth and the moon is about 20.46 x 10 newtons. Use these values along with the mass of Earth and the gravitational constant you found in part c, to estimate the mass of the moon. Type your response here:

Answer 1

Solution:

(a). The representative equation is,

$F=G\frac{m_1m_2}{d^2}$ ------------------------------------------------ (1)

(b). Re-writing equation 1 in terms of gravitational constant, G.

$F\times d^2=Gm_1m_2$

$G=\frac{F\times d^2}{m_1m_2}$

$G=\frac{Fd^2}{m_1m_2}$ ------------------------------------------------------ (2)

(c). According to the problem,

$d=2\times 10^8 km=2\times 10^8\times 10^3m=2\times 10^{11} m$

$F=64.32\times 10^{22}N$

$m_1=6\times 10^{24}kg$ [mass of Earth]

$m_2=6.4\times 10^{23}kg$ [mass of Marsh]

Plugging all these quantities into eqn 2,

$G=\frac{64.32\times 10^{22}\times (2\times 10^{11})^2}{6\times 10^{24}\times 6.4\times 10^{23}}$

$G=\frac{64.32\times 10^{22}\times 4\times 10^{22}}{6\times 10^{24}\times 6.4\times 10^{23}}$

$G=6.7\times 10^{-3}$  $Nm^2/kg^2$

[NOTE: the actual value of G is 6.7*10^-11 Nm²/Kg². But here it comes out of the power of -3. The difference could be due to the numerical values in the problem.]

(d). Re-writing eqn 2 in terms of m1,

$F=G\frac{m_1m_2}{d^2}$

$Fd^2=Gm_1m_2$

$m_1=\frac{Fd^2}{Gm_2}$

This is the required equation.