What is the value of the composite constant (gmer2e), to be multiplied by the mass of the object mo in the equation above? expre

Question

Gekata [30.6K]

2 years ago

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What is the value of the composite constant (gmer2e), to be multiplied by the mass of the object mo in the equation above? expre

ss your answer numerically in meters per second per second?

Physics

2 answers:

Bezzdna [24] · Answer 1 · 2023-10-23T06:14:24+03:00

The solution would be like this for this specific problem:

F = (G Me Mo) / Re^2

F / Mo = (G Me) / Re^2

G = gravitational constant = 6.67384 * 10^-11 m3 kg-1 s-2

Me = 5.972 * 10^24 kg

Re^2 = (6.38 * 10^6)^2 m^2 = 40.7044 * 10^12 m^2 = 4.07044 * 10^13 m^2

G Me / Re^2 = (6.67384 * 10-11 * 5.972 * 10^24) / 4.0704 * 10^13 = 9.7196 m/s^2

9.7196 m/s^2 = acceleration due to Earth’s gravity

Therefore, the value of the composite constant (Gme / r^2e) that is to be multiplied by the mass of the object mo in the equation above is 9.7196 m/s^2.

OverLord2011 [107] · Answer 2 · 2023-10-23T08:11:03+03:00

The value of the composite constant is to be multiplied by the mass of object is $\boxed{9.819\text{ m/s}^2}$ .

Further explanation:

We have to find the composite constant.

From the Newton’s law of the gravitation, gravitational force exerted by earth on the object at the surface of the earth can be calculated as,

$\boxed{F=\dfrac{{G{M_e}{m_o}}}{{{R_e}^2}}}$

Here, $G$ is the gravitational constant and its value is $6.674 \times10^{-11}\text{ m}^3/\text{kg}\cdot\text{s}^2$ .

${M_e}$ is the mass of the Earth which is $5.972\times10^{24}\text{ kg}$ .

${m_o}$ is the mass of object on surface of the Earth in kg

${R_e}$ is the distance between the center of Earth to the center of the object that is the radius of the Earth which is equal to $6.371 \times {10^6}\,{\text{m}}$ .

So, the gravitational force exerted by earth on the object of unit mass at the surface of the earth can be calculated as,

${F_1}=\dfrac{{G{M_e}}}{{{R_e}^2}}$

Substitute the value of $G$ as $6.674 \times10^{-11}\text{ m}^3/\text{kg}\cdot\text{s}^2$ , value of ${M_e}$ as $5.972\times10^{24}\text{ kg}$ and value of ${R_e}$ as $6.371\times{10^6}\text{m}$ in above equation.

$\begin{aligned}{F_1}&=\frac{{\left( {6.674 \times {{10}^{ - 11}}} \right)\left( {5.972 \times {{10}^{24}}} \right)}}{{{{\left( {6.371 \times {{10}^6}} \right)}^2}}}\\&=\frac{{\left( {3.9857 \times {{10}^{14}}} \right)}}{{40.59 \times {{10}^{12}}}}\\&=9.819\text{ m/s}^2\\\end{aligned}$