Answer:
F' = 73.7 N
F = 8.749×10⁻³ N
a' = a = 9.83 m/s²
Explanation:
(a)
For the rock
Applying
F' = Gm'm/r²................... Equation 1
Where F = magnitude of the gravitational force on the rock, G = Gravitational constant, m' = mass of the rock, m = mass of the earth, r = radius of the earth.
From the question,
Given: m' = 7.5 kg
Constant: m = 5.98×10²⁴ kg, G = 6.67×10⁻¹¹ Nm²/kg², r = 6.37×10⁶ m
Substitute these values into equation 1
F' = 6.67×10⁻¹¹ (7.5)(5.98×10²⁴)/(6.37×10⁶)²
F' = 7.37×10¹ N
F' = 73.7 N
Also, For the pebble,
F = GMm/r².............. Equation 2
Where M = mass of the pebble, F = Gravitational force exerted on the pebble by the earth
Given: M = 8.9×10⁻⁴ kg,
Substitute into equation 2
F = 6.67×10⁻¹¹(8.9×10⁻⁴)(5.98×10²⁴)/(6.37×10⁶)²
F = 8.749×10⁻³ N
(b)
For the rock,
a' = F'/m'
Where a' = magnitude of the acceleration of the rock
a' = 73.7/7.5
a' = 9.83 m/s²
For the pebble,
a = F/M
Where a = acceleration of the pebble
a = (8.749×10⁻³)/(8.9×10⁻⁴)
a = 9.83 m/s²
