Question

An object of mass m is dropped from height h above a planet of mass M and radius R .Find an expression for the object's speed as it hits the ground.Express your answer in terms of the variables m,M,h,R and appropriate constants.v= _____

Answer 1

An expression for the object's speed as it hits the ground is:

v = √ [ ( 2GMh ) / ( R ( R + h ) ) ]

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Further explanation

Let's recall the Gravitational Force formula:

$\boxed {F = G\ \frac{m_1 m_2}{R^2}}$

where:

F = Gravitational Force ( N )

G = Gravitational Constant ( = 6.67 × 10⁻¹¹ Nm²/kg² )

m = mass of object ( kg )

R = distance between object ( m )

Let us now tackle the problem!

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Given:

mass of object = m

height position of object = h

mass of planet = M

radius of planet = R

initial speed of object = u = 0 m/s

Asked:

final speed of object = v = ?

Solution:

We will calculate the object's speed by using Conservation of Energy formula as follows:

$Ep_1 + Ek_1 = Ep_2 + Ek_2$

$-G \frac{Mm}{R + h} + \frac{1}{2}m u^2 = -G \frac{Mm}{R} + \frac{1}{2}m v^2$

$-G \frac{Mm}{R + h} + \frac{1}{2}m (0)^2 = -G \frac{Mm}{R} + \frac{1}{2}m v^2$

$-G \frac{Mm}{R + h} = -G \frac{Mm}{R} + \frac{1}{2}m v^2$

$G \frac{Mm}{R} -G \frac{Mm}{R + h} = \frac{1}{2}m v^2$

$G \frac{M}{R} -G \frac{M}{R + h} = \frac{1}{2} v^2$

$v^2 = 2GM ( \frac{1}{R} -\frac{1}{R + h} )$

$v^2 = 2GM\frac{h}{R(R +h) }$

$\boxed {v = \sqrt { \frac{ 2GMh } { R(R +h) } } }$

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Learn more

• Unit of G : brainly.com/question/1724648
• Velocity of Runner : brainly.com/question/3813437

• Kinetic Energy : brainly.com/question/692781

• Acceleration : brainly.com/question/2283922

• The Speed of Car : brainly.com/question/568302

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Answer details

Grade: High School

Subject: Mathematics

Chapter: Gravitational Force

Answer 2

Answer:

Explanation:

Assuming that h is much smaller than R, then we can say the acceleration of gravity is approximately constant.

Potential energy = Kinetic energy

mgh = 1/2 mv²

v = √(2gh)

v = √(2 (MG/R²) h)

v = √(2 MGh) / R