Question

Work in a gravitational field For large distances from the surface of Earth, the gravitational force is given by F(x) = GMm / (x+R)², where G = 6.7×10^−11 N m²/kg² is the gravitational constant, M = 6×10^24 kg is the mass of Earth, m is the mass of the object in the gravitational field, R = 6.378×10⁶ m is the radius of Earth, and x≥0 is the distance above the surface of Earth (in meters).

a. How much work is required to launch a rocket with a mass of 500 kg in a vertical flight path to a height of 2500 km (from Earth’s surface)?

Accepted Answer

Identify the force function given by the problem: $$F(x) = \frac{GMm}{(x+R)^2$}$$, where $$x is $$the height above Earth's surface, $$G is $$the gravitational constant, $$M is $$Earth's mass, $$m is $$the rocket's mass, and $$R is $$Earth's radius. Recognize that work done against a variable force over a distance is calculated by integrating the force over that distance. Since the force depends on $$x$$, set up the integral for work as $$W = \int_{0}^{h} F(x) \, dx$$, where $$h is $$the height to which the rocket is launched (converted to meters). Substitute the expression for $$F(x)$$ into the integral: $$W = \int_{0}^{h} \frac{GMm}{(x+R)^2$} \, dx. $$Evaluate the integral by recognizing that $$\int \frac{1}{(x+R)^2$} \, dx = -\frac{1}{x+R} + C. $$Apply the limits of integration from $$0 to$$ $$h to $$find $$W = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right). $$Plug in the known values for $$G$$, $$M$$, $$m$$, $$R$$, and $$h ($$remembering to convert $$h = 2500 km to $$meters) into the expression for $$W to $$find the work required to launch the rocket to the specified height.

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Key Concepts

Gravitational Force and Potential Energy

Work Done by a Variable Force

Integration in Calculus