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

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

(A) 719 N

Explanation:

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

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;

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.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft

answer a fan shaped mass of sediment deposited at the base of a mountain front do to an abrupt change in stream gradient;