Question

Determine the value of the gravitational field at the surface of Earth if the gravitational field around an object of mass m is given by Fg / m (Fg is the gravitational force on the object due to Earth).

Answer 1

The gravitational field at the surface of Earth is calculated using Newton's law of universal gravitation. By equating an object's weight to the gravitational force and substituting values for Earth's mass and radius, we find the acceleration due to gravity is about 9.80 m/s².

Step-by-step explanation:

To determine the value of the gravitational field at the surface of Earth, we use Newton's law of universal gravitation, which correlates the gravitational force (F) to the masses involved (m and M) and the distance (r) between their centers. Specifically, F is equal to the gravitational constant (G) multiplied by the product of the two masses and divided by the square of the distance between their centers: F = G(mM)/r2. The weight (mg) of an object can be equated to the gravitational force (F) it experiences due to Earth's mass.

Cancelling the mass m from both sides of the equation and plugging in the known values for G, M (Earth's mass), and r (Earth's radius), we get the acceleration due to gravity (g) at Earth's surface. The values we use are G = 6.674 × 10-11 N·m2/kg2, M = 5.97 × 1024 kg, and r = 6.38 × 106 m.

When we solve for g, we find that it is approximately 9.80 m/s2, which is the acceleration due to gravity at the surface of Earth. Thus, the gravitational field strength g is the resulting force (weight) experienced by any mass we place at that point, directed toward the center of Earth.

Answer 2

Explanation:

The gravitational force on an object of mass m at a distance R from the earth's centre is given by

`F_g=G(mM_E)/(R^2)`

The gravitational field at a point at a distance R away from the centre of the earth is given by

`\vec{F}=(F_g)/(m)`

which in our case gives

`\vec{F}=(F_g)/(m)=G(mM_E)/(R^2)\cdot(1)/(m)`
`\boxed{\vec{F}=G(M_E)/(R^2)\text{.}}`

Now at the surface of the earth R = 6371 *1000 m and

G = 6.67 * 10^-11

M_E = 5.97 * 10^24 kg

Therefore, the above formula gives

`\vec{F}=(6.67\cdot10^(-11))((5.97\cdot10^(24)))/((6371\cdot1000)^2)`

which upon evaluating gives us

`\boxed{\vec{F}=9.81N/kg\text{.}}`

which is our answer!