Question

The planet Venus has a mass of 4.87 x 10^24 kg and a radius of 6.05 x 10^6 km. What is

the magnitude of the gravitational force that an 81-kg person would experience while
standing on the surface of Venus?
(A) 179 N
(B) 719 N
(C) 259 N
(D) 7.19 x 10^8N
(E) 4.19 x 10^9N

Answer 1

The gravitational force that an 81-kg person would experience while standing on the surface of Venus is approximately;

(A) 719 N

Step-by-step explanation:

The question relates to gravity calculation for which we use the Newton's gravitational law which is presented as follows;

`F=G \cdot (M \cdot m)/(r^(2))`

Where;

G = The universal gravitational constant = 6.67408 × 10⁻¹¹ m³kg⁻¹s⁻²

F = The gravitational force between the two masses, "M" and "m"

r = The distance between the centers of the two masses

M = The mass of one body interacting in the gravitational force

m = The mass of the other body interacting in the gravitational force

The given parameters are;

The mass of Venus, M = 4.87 × 10²⁴ kg

The radius of the planet Venus, r (is corrected) = 6.05 × 10³ km = 6.05 × 10⁶ km

The mass of the person standing on the surface of Venus, m = 81-kg

Therefore, by plugging in the values, we get;

`F=6.67408 * 10^(-11) \cdot (4.87 * 10^(24) * 81)/((6.05 * 10^6)^(2)) = 719.2744587391571613960795027662 \ N`

From Newton's third law of motion, the gravitational force that the 81-kg person experiences while standing on the surface of Venus = F ≈ 719 N.